Answer: $\langle 3, -1, -2 \rangle$

Use the cross product of normals: Normal vectors: $\mathbf{n}_1 = \langle 1, 1, 1 \rangle$ and $\mathbf{n}_2 = \langle 1, -1, 2 \rangle$. Compute the cross product: $\mathbf{n}_1 \times \mathbf{n}_2 = \langle (1)(2) - (1)(-1),\; (1)(1) - (1)(2),\; (1)(-1) - (1)(1) \rangle = \langle 3, -1, -2 \rangle$. Why the correct answer works: Option A is the computed cross product, which gives the direction of the line of intersection. Why distractors fail: Options B, C, and D result from various sign or arithmetic errors in the cross product computation.

Answer: Yes, the calculation is correct

Rewrite the plane in standard form: The plane $3x + 4y + 0z = 10$ has normal $\langle 3, 4, 0 \rangle$ and $d = -10$ in the form $3x + 4y + 0z - 10 = 0$. Apply the distance formula: $D = \frac{|3(0) + 4(0) + 0(0) - 10|}{\sqrt{9 + 16 + 0}} = \frac{10}{5} = 2$. The student's calculation is correct. Why distractors fail: Option A is wrong because a missing explicit $z$-term simply means $c = 0$, which is fine. Option C is wrong because the formula specifically measures distance from a point not on the plane. Option D incorrectly includes the constant in the denominator.

Answer: $x = 2 + 4t,\; y = -1,\; z = 3 - 2t$

Apply the parametric formula: From $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$, each component gives $x = 2 + 4t$, $y = -1 + 0t$, $z = 3 + (-2)t$. Why the correct answer works: Option B correctly uses the given point and direction vector: $x = 2 + 4t$, $y = -1$, and $z = 3 - 2t$. Why distractors fail: Option A swaps the roles of the point and direction components. Option C incorrectly adds a $t$ term to the $y$-component when the direction's $y$-component is $0$. Option D negates the direction vector.

Answer: $-4x - 7y + 5z = -9$

Find two vectors in the plane: The line's direction is $\mathbf{v} = \langle 2, 1, 3 \rangle$. The vector from the line's base point $(1,0,-1)$ to $P = (0,2,1)$ is $\overrightarrow{QP} = \langle -1, 2, 2 \rangle$. Compute the normal vector: $\mathbf{n} = \mathbf{v} \times \overrightarrow{QP} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ -1 & 2 & 2 \end{vmatrix} = \langle (1)(2)-(3)(2),\; (3)(-1)-(2)(2),\; (2)(2)-(1)(-1) \rangle = \langle -4, -7, 5 \rangle$. Write the plane equation: Using point $P = (0, 2, 1)$: $-4(x-0) - 7(y-2) + 5(z-1) = 0 \Rightarrow -4x - 7y + 14 + 5z - 5 = 0 \Rightarrow -4x - 7y + 5z = -9$. Verify: Check with the line's base point $(1, 0, -1)$: $-4(1) - 7(0) + 5(-1) = -4 - 5 = -9$ ✓. Check with $P = (0,2,1)$: $0 - 14 + 5 = -9$ ✓. Why distractors fail: Option A is the negation of the correct normal ($4x+7y-5z=9$) which represents the same plane, but the constant is wrong here — actually $4x+7y-5z=9$ is equivalent to $-4x-7y+5z=-9$, so Option A is the same plane written differently. Options B and C use incorrect normal vectors resulting from cross product computation errors.

Answer: Both are correct — the distance is $4$, and the sign indicates which side of the plane the point is on

Signed vs. unsigned distance: The signed distance $\frac{ax_0 + by_0 + cz_0 + d}{\sqrt{a^2 + b^2 + c^2}}$ can be positive or negative. The actual distance is the absolute value, but the sign indicates which side of the plane the point lies on relative to the normal. Why the correct answer works: Option C correctly synthesizes both observations: the distance is indeed $4$ (unsigned), and the negative sign tells us the point is on the side opposite to the normal direction. Why distractors fail: Option A dismisses the sign's geometric meaning. Option B incorrectly states the distance is $-4$ (distance is non-negative). Option D is wrong because the formula is valid.

Answer: The direction vector of the line

Recall the parametric line equation: In $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$, the vector $\mathbf{r}_0$ is a fixed point on the line and $\mathbf{v}$ is scaled by the parameter $t$ to generate all other points. Why the correct answer works: $\mathbf{v}$ determines the direction in which the line extends. As $t$ varies, the line moves parallel to $\mathbf{v}$, making it the direction vector. Why distractors fail: Option A is wrong because a normal vector is perpendicular to a surface, not along a line. Option C is wrong because $\mathbf{r}_0$ is a specific point, not the midpoint. Option D is wrong because $\mathbf{v}$ need not have unit length.

Answer: $\frac{x - 2}{3} = \frac{y + 1}{1} = \frac{z - 4}{-2}$

Eliminate the parameter: From the parametric equations, solve for $t$: $t = \frac{x-2}{3} = \frac{y+1}{1} = \frac{z-4}{-2}$. Why the correct answer works: Option B correctly reflects the point $(2, -1, 4)$ and direction $\langle 3, 1, -2 \rangle$, preserving the negative sign on $-2$. Why distractors fail: Option A drops the negative sign on the $z$-denominator. Option C uses incorrect signs on the point coordinates. Option D swaps the point and direction components.

Answer: $D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$

State the distance formula: The point-to-plane distance formula is $D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$. Why the correct answer works: Option B matches the standard formula exactly, with the absolute value in the numerator and the magnitude of the normal vector $\sqrt{a^2 + b^2 + c^2}$ in the denominator. Why distractors fail: Option A divides by the square of the normal's magnitude instead of its magnitude. Option C omits the absolute value, which could yield a negative distance. Option D does not correspond to any standard distance formula.

Answer: They are parallel but not identical

Compare direction vectors: $\langle 2, -2, 4 \rangle = 2\langle 1, -1, 2 \rangle$, so the direction vectors are parallel. Check if one point lies on the other line: Check if $(0, 1, 1)$ lies on $\mathbf{r}_1$: $1 + t = 0 \Rightarrow t = -1$. Then $y = 2 - (-1) = 3 \neq 1$. So the point is not on $\mathbf{r}_1$. Conclusion: Since the lines are parallel but do not share a point, they are parallel and distinct. Why distractors fail: Option A is wrong because parallel lines either coincide or never meet. Option B is wrong because the lines do not share a common point. Option D is wrong because skew lines have non-parallel direction vectors.

Answer: They are identical

Compare direction vectors: $\langle 4, 2, 6 \rangle = 2\langle 2, 1, 3 \rangle$, so the lines are parallel. Check if a point from one line lies on the other: Check $(3, 0, 3)$ on $\mathbf{r}_1$: $1 + 2t = 3 \Rightarrow t = 1$. Then $y = -1 + 1 = 0$ ✓ and $z = 0 + 3 = 3$ ✓. The point lies on $\mathbf{r}_1$. Conclusion: Since the lines are parallel and share a point, they are identical. Why distractors fail: Option A is wrong because the direction vectors are parallel. Option C is wrong because parallel lines cannot intersect at just one point. Option D is wrong because the lines do share a point.

Answer: $6x + 3y + 2z = 6$

Find two vectors in the plane: $\overrightarrow{AB} = \langle -1, 2, 0 \rangle$ and $\overrightarrow{AC} = \langle -1, 0, 3 \rangle$. Compute the normal via cross product: $\overrightarrow{AB} \times \overrightarrow{AC} = \langle (2)(3) - (0)(0),\; (0)(-1) - (-1)(3),\; (-1)(0) - (2)(-1) \rangle = \langle 6, 3, 2 \rangle$. Write the plane equation: Using point $A = (1,0,0)$: $6(x-1) + 3(y-0) + 2(z-0) = 0 \Rightarrow 6x + 3y + 2z = 6$. Why distractors fail: Option A gives the intercept form but with wrong coefficients. Options C and D have the normal vector components incorrectly assigned.

Answer: $\frac{5}{3}$

Apply the distance formula: $D = \frac{|2(3) - 2(-1) + 1(2) - 5|}{\sqrt{2^2 + (-2)^2 + 1^2}}$. Compute numerator and denominator: Numerator: $|6 + 2 + 2 - 5| = |5| = 5$. Denominator: $\sqrt{4 + 4 + 1} = \sqrt{9} = 3$. Result: $D = \frac{5}{3}$. Option C is correct. Why distractors fail: Option A likely results from an arithmetic error. Option B uses only the denominator. Option D uses only the numerator and forgets to divide.

Answer: Because $t$ can take any real value, including negative values, extending the line in both directions from $\mathbf{r}_0$

Role of the parameter $t$: In the parametric equation, $t \in (-\infty, \infty)$. Positive $t$ moves in the direction of $\mathbf{v}$, and negative $t$ moves in the opposite direction. Why the correct answer works: Option B correctly notes that allowing all real $t$ extends the line infinitely in both directions from $\mathbf{r}_0$. Why distractors fail: Option A restricts $t$ to positive values, which would only produce a ray. Option C is incorrect — the direction vector does not reverse. Option D is wrong because the single vector $\mathbf{v}$ with negative $t$ already handles the opposite direction.

Answer: $2x + y + 2z = 5$

Find two vectors in the plane: Direction of line: $\mathbf{v} = \langle 1, 0, -1 \rangle$. Vector from the base point $(2,1,0)$ to $P = (0,3,1)$: $\mathbf{w} = \langle -2, 2, 1 \rangle$. Compute the normal vector: $\mathbf{n} = \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -1 \\ -2 & 2 & 1 \end{vmatrix} = \langle (0)(1)-(-1)(2),\; (-1)(-2)-(1)(1),\; (1)(2)-(0)(-2) \rangle = \langle 2, 1, 2 \rangle$. Write the plane equation: Using the base point $(2, 1, 0)$: $2(x-2) + 1(y-1) + 2(z-0) = 0 \Rightarrow 2x + y + 2z = 5$. Verify: Check with $P = (0,3,1)$: $2(0) + 3 + 2(1) = 5$ ✓. Check with base point $(2,1,0)$: $4 + 1 + 0 = 5$ ✓. Why distractors fail: Option A ($2x+3y+2z=7$) uses an incorrect $y$-coefficient, resulting from an error in the cross product. Option B ($2x-y+2z=1$) has the wrong sign on $y$ and wrong constant. Option D ($-2x-y-2z=-7$) simplifies to $2x+y+2z=7$, which has the wrong constant.

Answer: The direction vector of the line of intersection

Key principle: The line of intersection lies in both planes, so it must be perpendicular to both normal vectors simultaneously. Why the correct answer works: The cross product of two vectors produces a vector perpendicular to both. Since the line must be perpendicular to both normals, the cross product gives the direction of that line. Why distractors fail: Option A is wrong because the cross product is a vector, not a point. Option C is wrong because 'normal to a line' is not a standard concept in 3D. Option D describes a geometric quantity unrelated to the cross product of normals.

Answer: Parallel but not identical

Compare normal vectors: Plane 1 has normal $\langle 1, 2, -1 \rangle$ and Plane 2 has normal $\langle 2, 4, -2 \rangle = 2\langle 1, 2, -1 \rangle$. The normals are parallel. Check if identical: Dividing Plane 2 by 2 gives $x + 2y - z = 3.5$, which differs from $x + 2y - z = 3$. So they are parallel but distinct. Why distractors fail: Option A is wrong because the constant terms differ after normalization. Option B is wrong because parallel normals mean parallel planes, not perpendicular. Option D is wrong because parallel planes do not intersect.

Answer: $P_1$ and $P_2$

Compare normal vectors: $P_1$ has normal $\langle 1,1,1\rangle$. $P_2$ has normal $\langle 2,2,2\rangle = 2\langle 1,1,1\rangle$. $P_3$ has normal $\langle 1,-1,1\rangle$. Identify parallel pair: $P_1$ and $P_2$ have proportional normals, so they are parallel. Dividing $P_2$ by 2 gives $x+y+z=2.5 \neq 1$, so they are parallel but distinct. Why distractors fail: Options A and B involve $P_3$, whose normal $\langle 1,-1,1\rangle$ is not proportional to $\langle 1,1,1\rangle$. Option D is wrong because $P_1$ and $P_2$ are indeed parallel.

Answer: It is perpendicular to every vector that lies in the plane

Definition of a normal vector: The vector $\langle a, b, c \rangle$ is the normal vector to the plane. By definition, it is orthogonal to any vector lying in the plane. Why the correct answer works: The plane equation states that $\langle a, b, c \rangle \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0$, meaning the normal is perpendicular to every displacement vector in the plane. Why distractors fail: Option A is wrong because the normal is perpendicular to the plane, not within it. Option C is wrong because a vector between two points on the plane lies in the plane. Option D is wrong because 'tangent to the plane' describes vectors in the plane, not the normal.

Answer: $3x - y + 4z = -3$

Apply the plane equation formula: Using $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$: $3(x-1) - 1(y-2) + 4(z+1) = 0$. Expand and simplify: $3x - 3 - y + 2 + 4z + 4 = 0 \Rightarrow 3x - y + 4z + 3 = 0 \Rightarrow 3x - y + 4z = -3$. Why distractors fail: Option A sets the right-hand side to $0$, forgetting to expand the point coordinates. Option C has the wrong sign on the constant. Option D results from an arithmetic error when combining the expanded terms.

Answer: Yes, any point on one parallel plane will give the same distance to the other plane

Why any point works: For parallel planes, every point on one plane is the same perpendicular distance from the other plane. The point-to-plane distance formula gives this perpendicular distance. Why the correct answer works: Option C correctly states that the choice of point on the first plane is irrelevant — the result is always the same for parallel planes. Why distractors fail: Option A is wrong because applying a point-to-plane formula with a point from another plane is perfectly valid. Option B is wrong because all points on a parallel plane are equidistant. Option D is wrong because the formula works regardless of algebraic form, as long as the equation is correct.

Answer: When two non-parallel planes intersect, they intersect along an entire line, not a single point

Identify the error: The student's claim that non-parallel planes meet in a single point is the mistake. In 3D, two distinct non-parallel planes always intersect along a line. Why the correct answer works: Solving two linear equations in three unknowns yields a one-parameter family of solutions — a line. Option B correctly identifies this. Why distractors fail: Option A is wrong because 'skew' applies to lines, not planes. Option C is wrong because the intersection of two planes (linear equations) is always linear. Option D is wrong because distinct parallel planes share no points.

Answer: $(1, 0, 1)$

Identify the condition for the $xz$-plane: The $xz$-plane is defined by $y = 0$. Solve for the parameter: From $y = -2 + t$, set $y = 0$: $t = 2$. Find the point: $x = 1 + 0(2) = 1$, $y = 0$, $z = 3 + (-1)(2) = 1$. The intersection point is $(1, 0, 1)$. Why distractors fail: Option B forgets to update $z$. Option C sets $z = 0$ instead of $y = 0$. Option D is wrong because $y$ does vary along the line.

Answer: $\langle -3, 5, 7 \rangle$

Set up the system with $z = t$: Substituting $z = t$: equation 1 becomes $x + 2y = 3 + t$ and equation 2 becomes $3x - y = 1 - 2t$. Solve the $2 \times 2$ system: Multiply equation 2 by 2 and add to equation 1: $x + 2y + 6x - 2y = 3 + t + 2 - 4t \Rightarrow 7x = 5 - 3t \Rightarrow x = \frac{5}{7} - \frac{3}{7}t$. Substituting back: $2y = 3 + t - x = 3 + t - \frac{5}{7} + \frac{3}{7}t = \frac{16}{7} + \frac{10}{7}t \Rightarrow y = \frac{8}{7} + \frac{5}{7}t$. Extract the direction vector: The parametric equations are $x = \frac{5}{7} - \frac{3}{7}t$, $y = \frac{8}{7} + \frac{5}{7}t$, $z = t$. The coefficients of $t$ are $\langle -\frac{3}{7}, \frac{5}{7}, 1 \rangle$, which is proportional to $\langle -3, 5, 7 \rangle$. Cross-check with the cross product method: $\mathbf{n}_1 \times \mathbf{n}_2 = \langle 1,2,-1 \rangle \times \langle 3,-1,2 \rangle = \langle (2)(2)-(-1)(-1),\; (-1)(3)-(1)(2),\; (1)(-1)-(2)(3) \rangle = \langle 3, -5, -7 \rangle$. This is the negation of $\langle -3, 5, 7 \rangle$, confirming both represent the same line direction. Why distractors fail: Option A ($\langle 3,-5,7\rangle$) incorrectly negates only two components. Option C ($\langle 3,-5,-7\rangle$) is the cross product result and represents the same direction as Option B, but since the question specifically asks for the direction obtained by the substitution method, Option B is the direct answer. Option D ($\langle -3,-5,7\rangle$) has an incorrect sign on the $y$-component.

Answer: The claim is incorrect because $\mathbf{n}_2 = -2\mathbf{n}_1$, so the planes are parallel and distinct, meaning they do not intersect

Check if normals are proportional: $\mathbf{n}_2 = -2\mathbf{n}_1$, so the normal vectors are parallel. This means the planes are parallel. Why the correct answer works: Since the planes are distinct (given) and have parallel normals, they are parallel but do not intersect. The claim is therefore incorrect. Why distractors fail: Option A is wrong because distinct parallel planes never intersect. Option C is wrong because $\mathbf{n}_1 \times \mathbf{n}_2 = \mathbf{0}$ when the vectors are parallel. Option D is wrong because the problem states the planes are distinct.

Answer: $\frac{6}{3} = 2$

Pick a point on one plane: On $2x - y + 2z = -2$, let $y = z = 0$: then $2x = -2$, so $x = -1$. The point is $(-1, 0, 0)$. Find distance to the other plane: $D = \frac{|2(-1) - 0 + 0 - 4|}{\sqrt{4 + 1 + 4}} = \frac{|-6|}{3} = 2$. Why distractors fail: Option A uses only the numerator without dividing. Option C uses an incorrect denominator. Option D results from using the difference of constants incorrectly.

Answer: $x = 1 + 2t,\; y = 4t,\; z = -2 + 4t$

Find the direction vector: The direction vector is $(3 - 1, 4 - 0, 2 - (-2)) = \langle 2, 4, 4 \rangle$. Write the parametric equations: Using the point $(1, 0, -2)$: $x = 1 + 2t$, $y = 0 + 4t$, $z = -2 + 4t$. Why distractors fail: Option B uses the wrong base point combined with the direction. Option C uses incorrect direction components (swapping point coordinates with direction). Option D omits the initial point entirely.

Answer: Incorrect; the $y$-coefficient is $0$, so the denominator should be $\sqrt{16 + 0 + 9} = 5$, and $D = \frac{|8 + 0 + 3 + 7|}{5} = \frac{18}{5}$

Identify the plane's coefficients: The plane $4x + 0y - 3z + 7 = 0$ has $a=4$, $b=0$, $c=-3$, $d=7$. Evaluate the denominator: $\sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{16+0+9} = 5$. The student's numerical answer for the denominator is correct, but they omitted $b = 0$ from the expression. Compute the distance: Numerator: $|4(2) + 0(3) + (-3)(-1) + 7| = |8 + 0 + 3 + 7| = 18$. So $D = \frac{18}{5}$. Why distractors fail: Option A says 'correct' but the student's denominator expression was incomplete (missing $b^2$). Option C is wrong because missing variables just have coefficient $0$. Option D uses $-3$ instead of $+3$ for the $cz_0$ term.