JEE Main 2024 — Three Dimensional Geometry Question with Solution

Let,

$L_1:\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}=\lambda$

$\Rightarrow M(3\lambda +1,2\lambda +2,-2\lambda -1)\equiv M\left(\alpha ,\beta ,\gamma \right)$

$\Rightarrow \alpha +\beta +\gamma =3\lambda +2$

Also, $L_2:\frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}=\mu$

$\Rightarrow N(-3\mu -2,-2\mu +2,4\mu +1)\equiv N\left(a,b,c\right)$

$\Rightarrow a+b+c=-\mu +1$

Now, from the above diagram, we can see $MN$ lie on the same line,

So, $\frac{3\lambda +2}{-3\mu -1}=\frac{2\lambda }{-2\mu }=\frac{-2\lambda -4}{4\mu -2}$

Now, solving $\frac{3\lambda +2}{-3\mu -1}=\frac{2\lambda }{-2\mu }$

$\Rightarrow 3\lambda \mu +2\mu =3\lambda \mu +\lambda$

$\Rightarrow 2\mu =\lambda ......\left(i\right)$

Now, solving $\frac{2\lambda }{-2\mu }=\frac{-2\lambda -4}{4\mu -2}$ we get,

$\Rightarrow 2\lambda \mu -\lambda =\lambda \mu +2\mu$

$\Rightarrow \lambda \mu =\lambda +2\mu$

$\Rightarrow \lambda \mu =2\lambda$ $\left\{\text{as} 2\mu =\lambda \right\}$

$\Rightarrow \mu =2 (\lambda \ne 0)$

$\Rightarrow \lambda =4$ $\left\{\text{from equation} \left(i\right)\right\}$

Hence, $\alpha +\beta +\gamma =14$ and $a+b+c=-1$

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