JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Three Dimensional Geometry

Points on $L_1$: $M=(3\lambda +1,2\lambda +2,-2\lambda -1)$, so $\alpha +\beta +\gamma =3\lambda +2$.

Points on $L_2$: $N=(-3\mu -2,-2\mu +2,4\mu +1)$, so $a+b+c=1-\mu$.

Collinearity of $(-1,2,3)$, $M$, $N$ means the vectors from $(-1,2,3)$ to each point are proportional:
$M-(-1,2,3)=(3\lambda +2,2\lambda ,-2\lambda -4)$, $N-(-1,2,3)=(-3\mu -1,-2\mu ,4\mu -2)$.

Equating the ratios of the first two components:
$\frac{3\lambda +2}{-3\mu -1}=\frac{2\lambda }{-2\mu }=-\frac{\lambda }{\mu }\Rightarrow \mu (3\lambda +2)=\lambda (3\mu +1)\Rightarrow 2\mu =\lambda$.

Equating the last two components' ratio to the same value:
$-\frac{\lambda }{\mu }=\frac{-2\lambda -4}{4\mu -2}\Rightarrow -\lambda (4\mu -2)=\mu (-2\lambda -4)\Rightarrow -4\lambda \mu +2\lambda =-2\lambda \mu -4\mu \Rightarrow -2\lambda \mu +2\lambda +4\mu =0\Rightarrow \lambda \mu =\lambda +2\mu$.

Substituting $\lambda =2\mu$: $2{\mu }^2=4\mu \Rightarrow \mu =2$ (rejecting $\mu =0$), so $\lambda =4$.

Then $\alpha +\beta +\gamma =3(4)+2=14$ and $a+b+c=1-2=-1$.

$\frac{(\alpha +\beta +\gamma {)}^2}{(a+b+c{)}^2}=\frac{196}{1}=196$.