Given: Equation of planes: x - 3y + z + 6 = 0 and x + 2y + 3z + 5 = 0
As we know that, the equation of a plane passing through the line of intersection of two planes
ax+by+cz+d=0 and
ax+by+cz+d=0 is given by:
(a1x+b1y+c1z+d1)+λ(a2x+b2y +c2z+d2)=0 where
λ ∈R ⇒(x−3y+z+6)+λ(x+2y +3z+5)=0 .........(1)
∵ It is given that plane passing through the line line of intersection of the plane x - 3y + z + 6 = 0 and x + 2y + 3z + 5 = 0 also passes through the point (0, 0, 0)
i.e The point (0, 0, 0) will satisfy the equation (1)
⇒(0+0+0+6)+λ(0+0 +0+5)=0 ⇒6+5λ=0 ⇒λ=−6∕5 Now by substituting λ = - 6/5 in equation (1) we get
⇒(x−3y+z+6)−.(x+2y +3z+5)=0 ⇒x+27y+13z=0 Hence, the equation of the required plane is: x + 27y + 13z = 0