Let L_1: frac{x-1}{3}=frac{y-1}{-1}=frac{z+1} | vedclass

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(D) For lines $L_1$ and $L_2$ to intersect at point $B$,the coordinates must satisfy: $(3\lambda+1, -\lambda+1, -1) = (2\mu+2, 0, \alpha\mu-4)$. From

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$216$

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(D) For lines $L_1$ and $L_2$ to intersect at point $B$,the coordinates must satisfy: $(3\lambda+1, -\lambda+1, -1) = (2\mu+2, 0, \alpha\mu-4)$. From the $y$-coordinate,$-\lambda+1 = 0 \implies \lambda = 1$. Substituting $\lambda=1$ into the $x$-coordinate: $3(1)+1 = 2\mu+2 \implies 4 = 2\mu+2 \implies \mu = 1$. Substituting $\mu=1$ into the $z$-coordinate: $-1 = \alpha(1)-4 \implies \alpha = 3$. Thus,point $B = (4, 0, -1)$. Line $L_2$ is given by $\frac{x-2}{2} = \frac{y}{0} = \frac{z+4}{3} = \delta$. So,any point $P$ on $L_2$ is $(2\delta+2, 0, 3\delta-4)$. The direction vector of $AP$ is $\vec{AP} = (2\delta+2-1, 0-1, 3\delta-4+1) = (2\delta+1, -1, 3\delta-3)$. Since $AP \perp L_2$,the dot product of $\vec{AP}$ and the direction vector of $L_2$ $(2, 0, 3)$ is zero: $2(2\delta+1) + 0(-1) + 3(3\delta-3) = 0 \implies 4\delta+2 + 9\delta-9 = 0 \implies 13\delta = 7 \implies \delta = \frac{7}{13}$. Point $P = (2(\frac{7}{13})+2, 0, 3(\frac{7}{13})-4) = (\frac{40}{13}, 0, -\frac{31}{13})$. $PB^2 = (4-\frac{40}{13})^2 + (0-0)^2 + (-1+\frac{31}{13})^2 = (\frac{12}{13})^2 + (\frac{18}{13})^2 = \frac{144+324}{169} = \frac{468}{169}$. Finally,$26\alpha(PB)^2 = 26 	imes 3 	imes \frac{468}{169} = 78 	imes \frac{36}{13} = 6 	imes 36 = 216$.

Let $L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in R$,be two lines,which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$,then the value of $26 \alpha(PB)^2$ is . . . . . . .

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