Question

Find the equation of plane passing through the intersection of the planes $2x+3y-z+1=0$ and $x+y-2z+3=0$ and perpendicular to the plane $3x-y-2z-4=0$ .

Answer 1

The $e{q}^n$ of plane after intersection :$(2x+3y-z+1)+\lambda (x+y-2x+3)=0$

$\Rightarrow x(2+\lambda )+y(3+\lambda )-z(1-2\lambda )+1+3\lambda =0$ __ (1)

This is $\perp$ to the plane $3x-y-2x-4=0$

[using :$a_1{a}_2+b_1{b}_2+c_1{c}_2=0]$

$3(2+\lambda )-1(3+\lambda )+2(1-2\lambda )=0$

$6+3\lambda -3-\lambda +2-4\lambda =0$ $\Rightarrow \lambda =\frac{5}{2}$

Putting $\lambda =\frac{5}{2}$ in eq (i)

$x(2+\frac{5}{2})+y(3+\frac{5}{2})-z(1-2\times \frac{5}{2})+1+3\times \frac{5}{2}=0$

$9x+11y+8z+17=0$