Question

Find the equation of the plane passing through the line of intersection of the planes $2x+y-z=3,5x-3y+4z+9=0$ and parallel to the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}$.

Answer 1

The equation of the plane passing through the line of intersection of the planes $2x+y-z=3$ and $5x-3y+4z+9=0$ is

$(2x+y-z-3)+\lambda (5x-3y+4z+9)=0$

$x(2+5\lambda )+y(1-3\lambda )+z(4\lambda -1)+9\lambda -3=0$

This plane is parallel to the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}$. (1)

Therefore,

$2(2+5\lambda )+4(1-3\lambda )+5(4\lambda -1)=0$

$18\lambda +3=0$

$\lambda =-\frac{1}{6}$

Putting the value of $\lambda$ in equation 1, we get,

$x(2-\frac{5}{6})+y(1+\frac{3}{6})+z(-\frac{4}{6}-1)-\frac{9}{6}-3=0$

$7x+9y-10z-27=0$

This is the equation of the required plane.