JEE Main 2022 — Three Dimensional Geometry Question with Solution

Given plane contain line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0=2\mathrm{x}-5\mathrm{y}-\mathrm{z}-3$

So equation of plane $A$

$4ax-y+5z-7a+\lambda \left(2x-5y-z-3\right)=0$ will satisfy $\left(4,-1,0\right)$

$\Rightarrow 16a+1-7a+\lambda \left(8+5-3\right)=0$

$\Rightarrow 9a+1+10\lambda =0 \cdots \left(1\right)$

Now normal vector of the plane $A$ is $\left(4a+2\lambda ,-1-5\lambda ,5-\lambda \right)$ and direction ratio along the line which contained the plane $A$ is $i-2j+k$

Since they will be perpendicular,

So, $4a+2\lambda +2+10\lambda +5-\lambda =0$

$\Rightarrow 11\lambda +4a+7=0 \cdots \left(2\right)$

Now on solving equation $\left(1\right)$ and $\left(2\right)$ we get $a=1,\lambda =-1$

So, now equation of plane will be $x+2y+3z-2=0$

Now let the point in the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}=t$ is $\left(7t+3,-t+2,-4t+3\right)$ satisfy the equation of plane $A$ $x+2y+3z-2=0$

$7t+3-2t+4+9-12t-2=0$

$\Rightarrow t=2$

So $\alpha +\beta +\gamma =7t+3-t+2-4t+3=2t+8=12$