The equation of the plane through the line of intersection of r (2 i -3 j +4 k )=1 and r ( i - j )+4=0 and perpandicular to r (2 i - j + k )+8=0 is r (-5 i +2 j +12 k )= . Then =

Step-by-step solution

Equation of plane through intersection: Given planes P₁: 2x - 3y + 4z - 1 = 0 and P₂: x - y + 4 = 0 , the family of planes through their intersection is (2x - 3y + 4z - 1) + (x - y + 4) = 0 . Simplifying gives (2+ )x + (-3- )y + 4z + (-1+4 ) = 0 with normal vector n = (2+ ) i + (-3- ) j + 4 k . This plane is perpendicular to P₃: 2x - y + z + 8 = 0 with normal n₃ = 2 i - j + k . Setting n n₃ = 0 yields: (2+ )(2) + (-3- )(-1) + 4 = 0 4 + 2 + 3 + + 4 = 0 3 = -11 = - 11 3 Substituting into the plane equation: (2 - 11 3