Find the equation of the plane that contans the line x1 y2t

Find the equation of the plane that contans the line x=1, y=2+t, z=-3t and the point p(2,0,-6)

Plane passes through ( 2, 0, -6)

Point (1, 2, 0) is also in plane. Setting t=1 in equiation of line we get another point

Point ( 1, 3, -3)

Let vector a from (1, 3, -3) to ( 1, 2, 0) = ( 0 , -1, 3)

Let vector b from ( 1, 3, -3) to ( 2, 0, -6) = ( 1,-3, -3)

Cross product of a x b = ( 12, 3, 1)

The general equation of a plane is: n.< x-x0 , y-yo , z-zo> =0

(12, 3,1).< x- 2, y-0, z+6 > =0

12(x- 2) +3y +1( z+6) =0

12x -24 +3y +z +6 =0

12x +3y +z -18 =0 Equation Of Plane