JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Three Dimensional Geometry

The direction vector of line $L$ is parallel to the cross product of the normals: $\vec{d}=(\hat{i}-\hat{j}+2\hat{k})\times (2\hat{i}+\hat{j}-\hat{k})=(-\hat{i}+5\hat{j}+3\hat{k})$.

To find a point on $L$, set $z=0$: from the plane equations we get $x-y=2$ and $2x+y=2$, giving $x=\frac{4}{3},y=-\frac{2}{3}$. So one point on $L$ is $A(\frac{4}{3},-\frac{2}{3},0)$.

The line $L$ can be written as $\vec{r}=(\frac{4}{3},-\frac{2}{3},0)+t(-1,5,3)$.

The foot of perpendicular $P$ from $(1,2,0)$ satisfies: $(P-(1,2,0))\cdot (-1,5,3)=0$.

Let $P=(\frac{4}{3}-t,-\frac{2}{3}+5t,3t)$. Then:
$(\frac{4}{3}-t-1,-\frac{2}{3}+5t-2,3t)\cdot (-1,5,3)=0$
$(\frac{1}{3}-t)(-1)+(-\frac{8}{3}+5t)(5)+3t(3)=0$
$-\frac{1}{3}+t-\frac{40}{3}+25t+9t=0$
$35t=\frac{41}{3}$, so $t=\frac{41}{105}$.

Then $\alpha +\beta +\gamma =\frac{4}{3}-\frac{41}{105}-\frac{2}{3}+5\cdot \frac{41}{105}+3\cdot \frac{41}{105}=\frac{2}{3}+7\cdot \frac{41}{105}=\frac{17}{5}$.

Thus $35(\alpha +\beta +\gamma )=35\cdot \frac{17}{5}=119$.