Question

Find the equation of the plane which passes through the line of intersection of the planes $2x+3y-4z+5=0$ and $x-5y+z+6=0$ and is parallel to the line $x+y-z+8=0=x-y-z+2$.

Answer 1

The equation of the plane passing through line of intersection of the planes is $2x+3y-4x+5=0$ and $x-5y+z+6=0$ is

$2x+3y-4x+5$$+a(x-5y+z+6)=0$ [Where $a$ being a scalar]

or, $x(2+a)+y(3-5a)+z(-4+a)+5+6a=0$.........(1)

Now, d.rs of the plane are $(2+a,3-5a,-4+a)$.

Given line (1) is parallel to the line represented by the planes $x+y-z+8=0=x-y-z+2$.......(2).

Now, the d.rs of the line represented by (2) be $l,m,n$ then we have the following relations

$l+m-n=0$ and $l-m-n=0$

or, $\frac{l}{0}=\frac{m}{0}=\frac{n}{-2}=k$(let) [$k\ne 0$ being a scalar]

$(l,m,n)=(0.0,-2k)$

Since (1) is parallel to (2) then the normal of (1) is perpendicular to (2).

Then,

$0(2+a)+0(3-5a)-2k(-4+a)=0$

or, $a=4$.

Then the plane is $6x-17y+0z+29=0$.