Question 6

Find the normal to plane $P$: The plane contains the line with direction vector $\vec{v} = (2, 3, 1)$ and is perpendicular to the plane $x + 2y + 3z = 4$ with normal $\vec{n}_Q = (1, 2, 3)$. The normal to $P$, $\vec{n}_P$, is $\vec{v} \times \vec{n}_Q = \det\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(9-2) - \hat{j}(6-1) + \hat{k}(4-3) = 7\hat{i} - 5\hat{j} + \hat{k}$.

Equation of $P$: Using point $(1, 3, -2)$ from the line, $7(x-1) - 5(y-3) + 1(z+2) = 0 \Rightarrow 7x - 5y + z + 10 = 0 \Rightarrow 7x - 5y + z = -10$. Thus, (A) is correct.

Equation of $P_1$: $P_1$ is parallel to $P$ and passes through $(4, 2, 2)$: $7(x-4) - 5(y-2) + (z-2) = 0 \Rightarrow 7x - 5y + z = 20$.

Distance between $P$ and $P_1$: $d = \frac{|20 - (-10)|}{\sqrt{7^2 + (-5)^2 + 1^2}} = \frac{30}{\sqrt{75}} = \frac{30}{5\sqrt{3}} = 2\sqrt{3}$. So this distance is not 30. Thus, (B) is incorrect.

Distance of $P$ from origin: $d_0 = \frac{|10|}{\sqrt{75}} = \frac{10}{5\sqrt{3}} = \frac{2}{\sqrt{3}}$. Thus, (C) is incorrect.

Angle with $2x + 2y + z = 3$: $\cos \theta = \frac{|(7, -5, 1) \cdot (2, 2, 1)|}{\sqrt{75}\sqrt{2^2+2^2+1^2}} = \frac{|14 - 10 + 1|}{5\sqrt{3} \cdot 3} = \frac{5}{15\sqrt{3}} = \frac{1}{3\sqrt{3}}$. Thus, (D) is correct.