JEE Main 2024 (Online) 30th January Evening ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
3D Geometry
Answer$196$
Solution
$M=(1+3t,2+2t,-1-2t)$, $N=(-2-3s,2-2s,1+4s)$, collinear with $(-1,2,3)$. Solving gives $t=4,s=2$: $M=(13,10,-9)$, $N=(-8,-2,9)$. $\alpha+\beta+\gamma=14$, $a+b+c=-1$. Ratio $=196$.