Three Dimensional Geometry

Concept:Find the equation of a plane containing a given line and perpendicular to another plane, then evaluate distances and angles for related planes.Explanation:The plane $P$ contains the line $\frac{x-1}{2}=\frac{y-3}{3}=\frac{z+2}{1}$ and is perpendicular to the plane $x+2y+3z=4$.The direction vector of the line is $(2,3,1)$ and the normal of the given plane is $(1,2,3)$.Since $P$ is perpendicular to that plane, its normal $\vec{n}$ must be perpendicular to both $(1,2,3)$ and $(2,3,1)$.Compute $\vec{n}$ using the cross product: $$\vec{n}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix}=7\hat{i}-5\hat{j}+\hat{k}.$$The line passes through $(1,3,-2)$, so $P$ passes through this point.Using point-normal form: $7(x-1)-5(y-3)+(z+2)=0$ simplifies to $7x-5y+z=-10$.Thus option A is correct.$P_1$ is parallel to $P$ and passes through $(4,2,2)$.Its equation is $7x-5y+z=d$. Substituting gives $d=20$, so $P_1:7x-5y+z=20$.Distance between parallel planes $7x-5y+z=-10$ and $7x-5y+z=20$ is $\frac{| -10-20 |}{\sqrt{7^2+(-5)^2+1^2}}=\frac{30}{\sqrt{75}}\neq30$.Thus option B is false.Distance from origin to $P$: $\frac{|0+0+0+10|}{\sqrt{75}}=\frac{10}{\sqrt{75}}=\frac{2}{\sqrt{3}}\neq2\sqrt3$.Hence option C is false.Acute angle between $P$ and $2x+2y+z=3$: normals $(7,-5,1)$ and $(2,2,1)$.$\cos\theta=\frac{|7(2)+(-5)(2)+1(1)|}{\sqrt{75}\cdot\sqrt{9}}=\frac{1}{3\sqrt3}$, so $\theta=\cos^{-1}\left(\frac{1}{3\sqrt3}\right)$.Option D is correct.Answer:Options A and D are true.