Question

Find the equation of the plane passing through the line of intersection of the planes $x+2y+z=3,2x-y-z=5$ and the point $(2,1,3)$. Also find direction rations of a normal to this plane.

Answer 1

The equation of the plane passing through the intersection of the planes.

$x+2y+z=3$ and $2x-y-z=5$ is given by

$(x+2y+z-3)+\lambda (2x-y-z)=5x(2\lambda +1)+y(2-\lambda )+z(1-\lambda )-3-5\lambda =0-\left(1\right)$

It passes through $(2,1,3)$ therefore,

$2(2\lambda +1)+(2-\lambda )+3(1-\lambda )-3-5\lambda =0\Rightarrow 4\lambda +2+2-\lambda +3-3\lambda -3-5\lambda =0\Rightarrow 4-5\lambda =0\Rightarrow \lambda =\frac{4}{5}$

Sustituting $\lambda =\frac{4}{5}$ in $\left(1\right)$ we get

$13x+6y+z-35=0$ as the equation of required plane clearly direction ratios if normal to this plane are proportional to $13,6,1$.