Let $L_1$ be the line of intersection of the planes given by the equations
$$2x + 3y + z = 4 \quad \text{and} \quad x + 2y + z = 5.$$
Let $L_2$ be the line passing through the point $P(2, -1, 3)$ and parallel to $L_1$. Let $M$ denote the plane given by the equation
$$2x + y - 2z = 6.$$
Suppose that the line $L_2$ meets the plane $M$ at the point $Q$. Let $R$ be the foot of the perpendicular drawn from $P$ to the plane $M$.
Then which of the following statements is (are) TRUE?
The length of the line segment $PQ$ is $9\sqrt{3}$
The length of the line segment $QR$ is 15
The area of $\triangle PQR$ is $\dfrac{3}{2}\sqrt{234}$
The acute angle between the line segments $PQ$ and $PR$ is $\cos^{-1}\!\left(\dfrac{1}{2\sqrt{3}}\right)$
Step-by-Step Solution
Key Concept: Direction of line of intersection via cross product of plane normals; foot of perpendicular formula
Direction of $L_1$: cross product of normals $(2,3,1)$ and $(1,2,1)$:
$\vec{d} = (3\cdot1-1\cdot2,\ 1\cdot1-2\cdot1,\ 2\cdot2-3\cdot1) = (1,-1,1)$.
$L_2$: $\mathbf{r} = (2,-1,3) + t(1,-1,1)$.
Intersection with $M: 2x+y-2z=6$:
$2(2+t)+(-1-t)-2(3+t) = 4+2t-1-t-6-2t = -3-t = 6$, so $t = -9$.
$Q = (2-9,\ -1+9,\ 3-9) = (-7, 8, -6)$.
$PQ = |t|\cdot|\vec{d}| = 9\sqrt{3}$. ✓ (A is TRUE)
Foot of perpendicular $R$ from $P(2,-1,3)$ to $M$: normal to $M$ is $\vec{n}=(2,1,-2)$, $|\vec{n}|=3$.
$PR = \dfrac{|2(2)+(-1)-2(3)-6|}{3} = \dfrac{|4-1-6-6|}{3} = \dfrac{9}{3} = 3$.
$QR = \sqrt{PQ^2 + PR^2 - 2\cdot PQ\cdot PR\cdot\cos\theta}$... actually use: $PR\perp$ plane $M$, $PQ$ in plane through $L_2$ and $P$.
Angle between $PQ$ and $PR$: $\cos\theta = \dfrac{\vec{d}\cdot\vec{n}}{|\vec{d}||\vec{n}|} = \dfrac{2-1-2}{\sqrt{3}\cdot3} = \dfrac{-1}{3\sqrt{3}}$.
So $QR = PQ\cdot\sin\theta$ where $\sin^2\theta = 1 - \dfrac{1}{27} = \dfrac{26}{27}$.
$QR = 9\sqrt{3}\cdot\sqrt{\dfrac{26}{27}} = 9\sqrt{3}\cdot\dfrac{\sqrt{26}}{3\sqrt{3}} = 3\sqrt{26}$. (B: 15 is FALSE)
Area of $\triangle PQR = \dfrac{1}{2}\cdot PQ\cdot PR\cdot\sin(\angle QPR)$. Since $PR\perp M$ and $Q$ on $M$, $\angle PQR = 90°$.
So area $= \dfrac{1}{2}\cdot QR\cdot PR = \dfrac{1}{2}\cdot3\sqrt{26}\cdot3 = \dfrac{9\sqrt{26}}{2}$.
Check option C: $\dfrac{3}{2}\sqrt{234} = \dfrac{3}{2}\cdot3\sqrt{26} = \dfrac{9\sqrt{26}}{2}$. ✓ (C is TRUE)
Correct Answer: A, C