Question

Find an equation of the plane.
The plane that passes through
(8, 0, −2)and contains the line
x = 6 − 2t, y = 3 + 5t, z = 3 + 2t

Answer 1

we have two points in that plane
(8, 0, -2) and (6, 3, 3)

a vector through those points has direction
(6-8)i + 3 j + (2+2)k
= -2 i + 3 j + 4 k
so we have two vectors parallel to plane, their cross product is normal to the plane
i j k
-2 5 2 the given line direction
-2 3 4 the line direction we found

i( 20-6) + j(-4+8) + k (-6+10)
= 14 i + 4 j + 4 k
so plane is
14 x + 4 y + 4 z = constant
but (8,0,-2) is in it so
112 -8 = constant
14 x + 4 y + 4 z = 104
check my arithmetic !