Question

equation of plane which passes through the point of intersection of lines

Answer 1

The equation of the plane which passes through the point of intersection of lines x−13=y−21=z−32 and x−31=y−12=z−23 and at greatest distance from the point (0,0,0) is

(A)4x+3y+5z=25(B)4x+3y+5z=50(C)3x+4y+5z=49(D)x+7y−5z=2

I found the point of intersection of two lines as (4,3,5).Let the equation of the required plane is ax+by+cz+d=0.Squared distance of plane ax+by+cz+d=0 from (0,0,0) is d2a2+b2+c2

We need to maximize d2a2+b2+c2 under the constraint 4a+3b+5c+d=0.

I cannot solve it further.