Question

The equation of the plane passing through $(1,-1,1)$ and through the line of intersection of the planes, $x+2y-3z+1=0$, and $3x-2y+4z+3=0$ is

• A. $3x-4y+2z=9=0$
• B. $2x-3y+z-6=0$
• C. $7x+6y-8z+7=0$
• D. $7x-6y-8z-5=0$

Answer 1

The correct option is C $7x+6y-8z+7=0$

Equation of the plane passing through the intersecting of planes

$x+2y-3z+1=0$ and $3x-2y+4z+3=0$ is

$\left(x+2y-3z+1\right)+\lambda \left(3x-2y+4z+3\right)=0-----\left(1\right)$

Since, it passing through the point $(1,-1,1)$

So,

$\left(1+2\left(-1\right)-3\left(1\right)+1\right)+\lambda \left(3-2\left(-1\right)+4+3\right)=0$

$\lambda =\frac{1}{4}$

Putting $\lambda =\frac{1}{4}$ in $\left(1\right)$

$\left(x+2y-3z+1\right)+\frac{1}{4}\left(3x-2y+4z+3\right)=0$

$7x+6y-8z+7=0$

This is the required equation of the plane.

Hence the option $C$ is correct.