Question

Find the equation of the plane passing through the intersection of the planes :
$x+y+z+1=0$ and $2x-3y+5z-2=0$ and the point $(-1,2,1)$.

Answer 1

The equations of the given planes are
$x+y+z+1=0$ ...$\left(i\right)$
$2x-3y+5z-2=0$ ...$\left(ii\right)$
The equation of any plane passing through the line of intersection of the planes $\left(i\right)$ and $\left(ii\right)$ is
$x+y+z+1+k(2x-3y+5z-2)=0$ ...$\left(iii\right)$
where $k$ is a parameter.

It passes through the point $(-1,2,1)$ if
$-1+2+1+1+k(2\times (-1)-3(2)+5\times 1-2)=0$
$\Rightarrow 3+k(-10+5)=0$
$\Rightarrow 3-5k=0$
$\Rightarrow 5k=3$
$\Rightarrow k=\frac{3}{5}$

Substituting this value of $k$ in (iii),
$\Rightarrow x+y+z+1+\frac{3}{5}(2x-3y+5z-2)=0$
$5x+5y+5z+5+6x+9y+15z-6=0$
$11x-4y+20z-1=0$
which is the equation of the required plane.