The equation of the plane passing through (-2,2,2) and (2,-2,-2) and perpendicular to the plane 9 x-13 y-3 z=0 is

Step-by-step solution

Equation of plane passing through (-2,2,2) is a(x+2)+b ( y -2)+ c ( z -2)=0 Since this plane also passes through (2,-2,-2) , we get 4 a -4 ~b -4 c =0 a - b - c =0 Normal of the plane is parallel to 9 x-13 y-3 z=0 9 a-13 b-3 c=0 Solving (1) and (2), we write a | array cc -1 & -1 -13 & -3 array | = b | array cc 1 & -1 0 & -3 array | = c | array cc 1 & -1 9 & -13 array | a -10 = - b 6 = c -4 a 5 = b 3 = c 2 Hence equation of required plane is 5(x+2)+3(y-2)+2(z-2)=0 i.e. 5 x+3 y+2 z=0