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(B) The given line is $L_1: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$. Let $P(1,3,4)$ be a point on $L_1$. The foot of the perpendicular from $P$ to

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$26$

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(B) The given line is $L_1: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$. Let $P(1,3,4)$ be a point on $L_1$. The foot of the perpendicular from $P$ to the plane $x-2y-z-3=0$ is $Q$. The line $PQ$ is $\frac{x-1}{1}=\frac{y-3}{-2}=\frac{z-4}{-1} = t$. So $Q = (t+1, -2t+3, -t+4)$. Since $Q$ lies on the plane,$(t+1) - 2(-2t+3) - (-t+4) = 3 \Rightarrow t+1+4t-6+t-4=3 \Rightarrow 6t=12 \Rightarrow t=2$. Thus,$Q = (3,-1,2)$.The intersection point $R$ of $L_1$ and the plane is found by $2k+1 - 2(k+3) - (2k+4) = 3 \Rightarrow 2k+1-2k-6-2k-4=3 \Rightarrow -2k=12 \Rightarrow k=-6$. So $R = (-11,-3,-8)$.The line $L$ passes through $Q(3,-1,2)$ and $R(-11,-3,-8)$. The direction vector of $L$ is $\vec{v} = Q-R = (14, 2, 10)$,which is parallel to $(7, 1, 5)$.The distance $d$ of point $A(0,0,6)$ from line $L$ passing through $B(3,-1,2)$ with direction $\vec{v} = (7,1,5)$ is given by $d = \frac{|\vec{AB} 	imes \vec{v}|}{|\vec{v}|}$.$\vec{AB} = (3-0, -1-0, 2-6) = (3, -1, -4)$.$\vec{AB} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & -4 \ 7 & 1 & 5 \end{vmatrix} = \hat{i}(-5+4) - \hat{j}(15+28) + \hat{k}(3+7) = (-1, -43, 10)$.$d^2 = \frac{(-1)^2 + (-43)^2 + 10^2}{7^2 + 1^2 + 5^2} = \frac{1 + 1849 + 100}{49 + 1 + 25} = \frac{1950}{75} = 26$.

Let the line $L$ be the projection of the line $\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$ in the plane $x-2y-z=3$. If $d$ is the distance of the point $(0,0,6)$ from $L$,then $d^2$ is equal to .... .

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