Let be the plane passing through the point (3,-3,1) and perpendicular to the line joining the points (3,4,-1) and (2,-1,5) . If the equation of the plane containing the points (3,4,-1),(-1,2,5) and perpendicular to the plane is a x+y+c z-d=0 , then 3(a+c)=

Step-by-step solution

Plane passes through (3,-3,1) and perpendicular to the line joining the points (3,4,-1) and (2,-1,5) . DR 's of normal to the plane are (3-2),(4,+1),(-1-5) (1,5,-6) Equation of plane is given by x+5 y-6 z+d=0 consists the point (3,-3,1) . array ll & 3-15-6+d=0 d=18 & x+5 y-6 z+18=0 array Equation of plane containing points (3,4,-1) and (-1,2,5) is a x+y+c z-d=0 ...(i) Normal to this plane will be perpendicular to the line joining the points (3,4,-1) and (-1,2,5) . Then, a(3+1)+1(4-2)+c(-1-5)=0 4 a+2-6 c=0 2 a+1-3 c