Question

Find the equation of the 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

Given equation of planes are $2x+3y-z+6=0$ and $x+y-2z+3=0$

As we know that

The equation of plane passing through the line of intersection of the planes $a_{1}x+b_{1}y+c_{1}z+d_1=0$ and $a_{2}x+b_{2}y+c_{2}z+d_2=0$ is $a_{1}x+b_{1}y+c_{1}z+d_1+\lambda (a_{2}x+b_{2}y+c_{2}z+d_2)=0$

So the required equation of plane is $2x+3y-z+1+\lambda (x+y-2z+3)=0$

$⟹(2+\lambda )x+(3+\lambda )y+(-1-2\lambda )z+1+3\lambda =0\cdots \left(1\right)$

This plane is perpendicular to $3x-y-2z-4=0$

$3(2+\lambda )-(3+\lambda )-2(1-2\lambda )=0$ $(\because a_1{a}_2+b_1{b}_2+c_1{c}_2=0)$

$⟹\lambda =-\frac{5}{6}$

Substituting the $\lambda$ value in $\left(1\right)$

$(2-\frac{5}{6})x+(3-\frac{5}{6})y+(-1-2(-\frac{5}{6}))z+1+3(\frac{-5}{6})=0$

$⟹7x+13y+4z-9=0$

$\therefore$ Equation of required plane is $7x+13y+4z-9=0$