JEE Main 2021 — Three Dimensional Geometry Question with Solution

Given, $x+2y+z-6=0 ...\left(1\right)$$y+2z-4=0 ...\left(2\right)$

On solving, equations $\left(1\right)$ and $\left(2\right)$, we get

$x-3z=-2\Rightarrow x=3z-2\Rightarrow y=4-2z$

Hence, Equation of the line is $$

$\frac{x+2}{3}=\frac{y-4}{-2}=\frac{z}{1}$

$\Rightarrow$ line of intersection of two planes is

$\frac{x+2}{3}=\frac{y-4}{-2}=\frac{z}{1}=\lambda$ (Let)

It is clear from the diagram, $AP{\perp }^{ar}$ to line 

$\therefore \overline{\mathrm{AP}}.\left(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)=0$

$\left(3\lambda -5\right).3+\left(-2\lambda +2\right)\left(-2\right)+\left(\lambda -1\right).1=0$

$9\lambda -15+4\lambda -4+\lambda -1=0$

$14\lambda =20$

$\lambda =\frac{10}{7}\Rightarrow P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)$

$\Rightarrow \alpha +\beta +\gamma =\frac{16+8+10}{7}=\frac{34}{7}$

$21\left(\alpha +\beta +\gamma \right)=102$