Question

The plane x+3y-13=0 passes through the line of intersection of planes 2x-8y+4z+p=0 and 3x-5y+4z-10=0. If the plane is perpendicular to the plane 3x-y-2z-4=0, then the value of p is

Answer 1

ANSWER

Equation of plane passes through line of intersection of the plane 2x−8y+4z=p and 3x−5y+4z+10=0 is

(2x−8y+4z−p)+λ(3x−5y+4z+10)=0

⇒(2+3λ)x+(−8−5λ)y+(4+4λ)z−p+10λ=0 ...(i)

Given equation of plane is

x+3y+13=0 ....(ii)

Equations (i) and (ii) represent same plane.

∴4+4λ=0⇒λ=−1

Putting λ=−1 in equation (i), we get

−x−3y−p−10=0

⇒x+3y+p+10=0 ...(iii)

Comparing the coefficient of x,y and z constants term for equations (i) and (ii), we get

p+10=13

⇒p=3