Question

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

Answer 1

Equation of the planes are

$2x+y-z=3$ and $5x-3y+4z+9=0$

The equation of the plane passing through the line of intersection of these planes is

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

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

The plane is parallel to the line

$\frac{x-1}{2}=\frac{y-3}{3}=\frac{z-5}{5}$

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

i.e., $18\lambda +3=0$

$\Rightarrow 18\lambda =-3$

$\Rightarrow \lambda =-3/186$

$\therefore \lambda =-1/6$

Put the value $\lambda$ in equation-$\left(1\right)$ we obtain

$x(2+5(-1/6\left)\right)+y(1-3(-1/6\left)\right)+z\left(4\right(-1/6)-1)+9(-1/6)-3=0$

$\Rightarrow x(2-5/6)+y(1+3/6)+z(-4/6-1)-(9/6-3)=0$

$\Rightarrow x(12-5/6)+y(6+3/6)+z(-4-6/6)-9-18/6=0$

$\Rightarrow x\cdot 7/6+y\left(9/6\right)+z(-10/6)-27/6=0$

$\Rightarrow 1/6(7x+9y-10z-27)=0$

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

$\therefore$ The required equation of the plane is $7x+9y-10z-27=0$.