(717, #31) Find an equation of the plane that passes through the point ( 1, 6, -4 ) and contains the line x = 1 + 2t, y = 2 - 3t, z = 3

(717, #31) Find an equation of the plane that passes through the point ( 1, 6, -4 ) and contains the line
x = 1 + 2t, y = 2 - 3t, z = 3 - t.
Solution:
The points ( 1, 6, -4 ) and ( 1, 2, 3 ) are on the plane. Setting t = 1, we get another point ( 3, -1, 2 ) which is also on the plane.
Let a be the vector from ( 3, -1, 2 ) to ( 1, 2, 3 ); a = < -2, 3, 1 >
Let b be the vector from ( 3, -1, 2 ) to ( 1, 6, -4 ); b = < -2, 7, -6 >
The normal of the two vectors is:

The general equation of a plane is:

or

or
-25( x - 1 ) - 14( y - 6 ) -8( z + 4 ) = 0
or
-25x - 14y -8z = -77

by: nl