Since the lines are both parallel to the plane their cross product is perpindicular to the plane, and therefore is the normal
N = (-4 i + 4j + 8k)x(3i + 5j + 8k)
For a point in the plane use (-7,-6,-3)
Find an equation for the plane that contains the line with parametric equations
x = -7 - 4 t
y = -6 + 4 t
z = -3 + 8 t
and is parallel to the line with parametric equations
x = 3 t
y = -7 + 5 t
z = 5 + 8 t
. Find the the normal vector and the EquationOfPlane.
The equation should involve x, y, and z --I have no idea what you are doing.
Once you have N = a i + bj + ck
Use a(x -x0) + b(y-y0) + c(z-z0) = 0
to generate the eqn of the plane where (x0,y0,z0) is the known point