Question

find the equation of the plane passing through the intersection of the planes 2x+3y-z+1=0,x+y-2z+3=0, and perpendicular to the plane 3x-y-2z-4=0

Answer 1

equation of unknown plane,
(2x +3y -z + 1) +¢(x + y -2z + 3) =0

(2 +¢)x + (3 + ¢ )y +( -1 -2¢)z + (1 +3¢) =0

this plane is perpendicular upon below plane ,
3x - y -2z -4 =0

so,
3(2 + ¢ ) -(3 + ¢ ) -2( -1 -2¢) =0

6 + 3¢ -3 -¢ +2 +4¢ = 0

5 + 6¢ = 0

¢ = -5/6

put ¢ = -5/6 in equation of plane ,
6(2x + 3y -z + 1) -5(x + y -2z +3) =0

12x +18y -6z +6 -5x -5y +10z-15=0

7x + 13y + 4z -9 =0