Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L:\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}$ and $R(5,p,q)$ be a point on $L$. Then the square of the area of $\triangle PQR$ is ________.
Step-by-Step Solution
Key Concept: Find $T$ (foot of perpendicular from $Q$ to $L$), compute $P=2T-Q$, locate $R$ on $L$ using its $x$-coordinate, then use $\text{Area}=\tfrac{1}{2}|PQ||RT|$ (since $PQ\perp L$ and $RT\parallel L$).
General point $T=(2\lambda+1,3\lambda-1,4\lambda)$ on $L$. $\overrightarrow{QT}\cdot(2,3,4)=0$.
$(2\lambda-6)\cdot2+(3\lambda+1)\cdot3+(4\lambda-5)\cdot4=0 \Rightarrow 4\lambda-12+9\lambda+3+16\lambda-20=0 \Rightarrow 29\lambda=29 \Rightarrow \lambda=1$.
$T=(3,2,4)$. $P=2T-Q=(-1,6,3)$.
$R$ on $L$: $2\lambda+1=5 \Rightarrow \lambda=2$, so $R=(5,5,8)$.
$QT=\sqrt{16+16+1}=\sqrt{33}$, $RT=\sqrt{4+9+16}=\sqrt{29}$, $PQ=2QT=2\sqrt{33}$.
$\text{Area}=\tfrac{1}{2}\cdot2\sqrt{33}\cdot\sqrt{29}=\sqrt{29\cdot33}=\sqrt{957}$.
$\text{Area}^2=29\times33=957$.
Correct Answer: 957