Question

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

Answer 1

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

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

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

Since this plane is parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ whose direction ratio are $2,4,5$.

$\therefore$ The normal to the plane is perpendicular to this line

$\therefore$ $(2+4\lambda )2+[-(1+3\lambda \left)\right]4+(1+5\lambda )5=0$

$\Rightarrow$ $4+8\lambda -4-12\lambda +5+25\lambda =0$

$\Rightarrow$ $5+21\lambda =0$

$\Rightarrow$ $\lambda =\frac{-5}{21}$

Substituting $\lambda =\frac{-5}{21}$ in equation (i), we get

$(2x-y+z-3)+\left(\frac{-5}{21}\right)(4x-3y+5z+9)=0$

$\Rightarrow 42x-21y+21z-63-20x+15y-25z-45=0$

$\Rightarrow 22x-6y-4z-108=0$

$\Rightarrow 2(11x-3y-2z-54)=0$

$\Rightarrow 11x-3y-2z-54=0$