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(B) Let the point $M$ on $L_1$ be $(3\lambda+1, 2\lambda+2, -2\lambda-1)$.Then $\alpha+\beta+\gamma = (3\lambda+1) + (2\lambda+2) + (-2\lambda-1) = 3\

Vedclass · Accepted Answer

$196$

Vedclass · Answer

(B) Let the point $M$ on $L_1$ be $(3\lambda+1, 2\lambda+2, -2\lambda-1)$.Then $\alpha+\beta+\gamma = (3\lambda+1) + (2\lambda+2) + (-2\lambda-1) = 3\lambda+2$.Let the point $N$ on $L_2$ be $(-3\mu-2, -2\mu+2, 4\mu+1)$.Then $a+b+c = (-3\mu-2) + (-2\mu+2) + (4\mu+1) = -\mu+1$.The line passes through $P(-1, 2, 3)$,$M$,and $N$. Thus,the vectors $\vec{PM}$ and $\vec{PN}$ are collinear.$\vec{PM} = (3\lambda+2, 2\lambda, -2\lambda-4)$ and $\vec{PN} = (-3\mu-1, -2\mu, 4\mu-2)$.Since they are collinear,$\frac{3\lambda+2}{-3\mu-1} = \frac{2\lambda}{-2\mu} = \frac{-2\lambda-4}{4\mu-2}$.From $\frac{2\lambda}{-2\mu} = \frac{3\lambda+2}{-3\mu-1}$,we get $\lambda(-3\mu-1) = -\mu(3\lambda+2) \Rightarrow -3\lambda\mu - \lambda = -3\lambda\mu - 2\mu \Rightarrow \lambda = 2\mu$.From $\frac{2\lambda}{-2\mu} = \frac{-2\lambda-4}{4\mu-2}$,we get $\frac{\lambda}{-\mu} = \frac{-\lambda-2}{2\mu-1} \Rightarrow 2\lambda\mu - \lambda = \lambda\mu + 2\mu \Rightarrow \lambda\mu = \lambda + 2\mu$.Substituting $\lambda = 2\mu$,we get $(2\mu)\mu = 2\mu + 2\mu \Rightarrow 2\mu^2 = 4\mu \Rightarrow \mu = 2$ (as $\mu 
eq 0$).Then $\lambda = 4$.Thus,$\alpha+\beta+\gamma = 3(4)+2 = 14$ and $a+b+c = -(2)+1 = -1$.Therefore,$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2} = \frac{14^2}{(-1)^2} = 196$.

Let a line passing through the point $(-1, 2, 3)$ intersect the lines $L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}$ at $M(\alpha, \beta, \gamma)$ and $L_2: \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}$ at $N(a, b, c)$. Then the value of $\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$ equals

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