JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Three Dimensional Geometry

The line $L$ is the intersection of the two planes $x+2y+z=6$ and $y+2z=4$. The direction vector of $L$ is perpendicular to both normal vectors $\overrightarrow{n_1}=(1,2,1)$ and $\overrightarrow{n_2}=(0,1,2)$. Thus $\vec{d}=\overrightarrow{n_1}\times \overrightarrow{n_2}=(3,-2,1)$. To find a point on $L$, set $z=0$: from $y+2z=4$ we get $y=4$, and from $x+2y+z=6$ we get $x=-2$. So $L$ passes through $(-2,4,0)$ with direction $(3,-2,1)$. The parametric form is $(x,y,z)=(-2,4,0)+t(3,-2,1)$. The foot of perpendicular $P$ from $(3,2,1)$ satisfies $(P-(3,2,1))\cdot (3,-2,1)=0$. Substituting $P=(-2+3t,4-2t,t)$, we get $(-5+3t,2-2t,t-1)\cdot (3,-2,1)=0$, which gives $3(-5+3t)-2(2-2t)+(t-1)=0$. Solving: $-15+9t-4+4t+t-1=0\Rightarrow 14t=20\Rightarrow t=\frac{10}{7}$. Thus $P=(-2+\frac{30}{7},4-\frac{20}{7},\frac{10}{7})=(\frac{16}{7},\frac{8}{7},\frac{10}{7})$. Therefore $\alpha +\beta +\gamma =\frac{34}{7}$ and $21(\alpha +\beta +\gamma )=21\cdot \frac{34}{7}=102$.