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(B) Let the line $L$ be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4} = \lambda$. Any point on $L$ is $(2\lambda+1, 3\lambda-1, 4\lambda)$.Since $R(5, p, q

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$957$

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(B) Let the line $L$ be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4} = \lambda$. Any point on $L$ is $(2\lambda+1, 3\lambda-1, 4\lambda)$.Since $R(5, p, q)$ lies on $L$,we have $2\lambda+1 = 5 \implies \lambda = 2$. Thus,$R = (5, 5, 8)$.Let $T$ be the foot of the perpendicular from $Q(7, -2, 5)$ to the line $L$. Let $T = (2\lambda+1, 3\lambda-1, 4\lambda)$.The vector $\vec{QT} = (2\lambda+1-7, 3\lambda-1+2, 4\lambda-5) = (2\lambda-6, 3\lambda+1, 4\lambda-5)$.Since $\vec{QT}$ is perpendicular to the direction vector of the line $\vec{b} = (2, 3, 4)$,we have $\vec{QT} \cdot \vec{b} = 0$.$2(2\lambda-6) + 3(3\lambda+1) + 4(4\lambda-5) = 0 \implies 4\lambda - 12 + 9\lambda + 3 + 16\lambda - 20 = 0 \implies 29\lambda = 29 \implies \lambda = 1$.Thus,$T = (3, 2, 4)$.$P$ is the image of $Q$ in $L$,so $T$ is the midpoint of $PQ$. Thus,$QT = TP$.The area of $	riangle PQR = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes RT 	imes (QT + TP) = \frac{1}{2} 	imes RT 	imes (2QT) = RT 	imes QT$.$QT = \sqrt{(3-7)^2 + (2+2)^2 + (4-5)^2} = \sqrt{(-4)^2 + 4^2 + (-1)^2} = \sqrt{16+16+1} = \sqrt{33}$.$RT = \sqrt{(5-3)^2 + (5-2)^2 + (8-4)^2} = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29}$.Area of $	riangle PQR = \sqrt{33} 	imes \sqrt{29} = \sqrt{957}$.Therefore,the square of the area is $(\sqrt{957})^2 = 957$.

Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is $\qquad$

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