JEE Main 2022 (Online) 27th July Morning ShiftQuestion 1
Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac z1$ and $4ax-y+5z-7a=0=2x-5y-z-3$, $a\in\mathbb{R}$ at the point $P(\alpha,\beta,\gamma)$. Then the value of $\alpha+\beta+\gamma$ equals ________.
3D Geometry
Answer$12$
Solution
Line 1: $\frac{x-4}{1}=\frac{y+1}{-2}=\frac z1$, direction $d_1=(1,-2,1)$. Line 2 is the intersection of $4ax-y+5z-7a=0$ and $2x-5y-z-3=0$, direction $d_2=(4a,-1,5)\times(2,-5,-1)$. A point on it is $z=0$: $(x,y)$ solving $4ax-y=7a,\ 2x-5y=3$. Coplanarity of the two lines gives the determinant $59-59a=0\Rightarrow a=1$. For $a=1$: $d_2=(26,14,-18)$, normal $=d_1\times d_2=(22,44,66)\parallel(1,2,3)$, plane through $(4,-1,0)$: $x+2y+3z=2$. Line: $(3+7t,\,2-t,\,3-4t)$: $3+7t+4-2t+9-12t=2\Rightarrow -7t=-14\Rightarrow t=2$. $P=(17,0,-5)$, so $\alpha+\beta+\gamma=12$.