Let Q and R be the feet of perpendiculars fro | vedclass

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(D) The line $L_1$ is given by $\frac{x}{1} = \frac{y}{1} = \frac{z-1}{0} = r$. Thus,$Q = (r, r, 1)$.Since $PQ \perp L_1$,the vector $\vec{PQ} = (r-a,

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$12$

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(D) The line $L_1$ is given by $\frac{x}{1} = \frac{y}{1} = \frac{z-1}{0} = r$. Thus,$Q = (r, r, 1)$.Since $PQ \perp L_1$,the vector $\vec{PQ} = (r-a, r-a, 1-a)$ is perpendicular to the direction vector $(1, 1, 0)$.$(r-a)(1) + (r-a)(1) + (1-a)(0) = 0 \implies 2(r-a) = 0 \implies r = a$. So,$Q = (a, a, 1)$.The line $L_2$ is given by $\frac{x}{1} = \frac{y}{-1} = \frac{z+1}{0} = k$. Thus,$R = (k, -k, -1)$.Since $PR \perp L_2$,the vector $\vec{PR} = (k-a, -k-a, -1-a)$ is perpendicular to the direction vector $(1, -1, 0)$.$(k-a)(1) + (-k-a)(-1) + (-1-a)(0) = 0 \implies k-a + k+a = 0 \implies 2k = 0 \implies k = 0$. So,$R = (0, 0, -1)$.Given $\angle QPR = 90^{\circ}$,we have $\vec{PQ} \cdot \vec{PR} = 0$.$\vec{PQ} = (a-a, a-a, 1-a) = (0, 0, 1-a)$.$\vec{PR} = (0-a, 0-a, -1-a) = (-a, -a, -1-a)$.$(0)(-a) + (0)(-a) + (1-a)(-1-a) = 0$.$-(1-a)(1+a) = 0 \implies -(1-a^2) = 0 \implies a^2 = 1$.Therefore,$12a^2 = 12(1) = 12$.

Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a, a, a)$ on the lines $x=y, z=1$ and $x=-y, z=-1$ respectively. If $\angle QPR$ is a right angle,then $12a^2$ is equal to

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