Does the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lie in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)? Justify your answer algebraically

Answer 1

The cross-product of the direction vectors on the given plane (1,1,-1) and (2,-1,1) is (0,3,3). ---- I assume you know a method to find that cross-product.

So the equation of the plane is 3y + 3z = k
with (3,0,2) on that plane, so
3(0) + 3(2) = k, k = 6 and the equation of the plane is
3y + 3z = 6 or
y + z = 2

let's see where
x = 5 + u
y = -4 + 4u
z = 6 - u
intersects that plane:
-4 + 4u + 6 - u = 2
3u = 0
u = 0
So the given line intersects the plane at the point (5,-4,6), so it cannot be part of the plane.
If the line is part of the plane, there would have to be an infinite number of solutions, that is, the above equation should have resulted in 0=0

another way:
To be on the plane, the direction vector of the line must be perpendicular to the normal of the plane.
We already found the normal to be (0,3,3) or (0,1,1) in reduced form
(0,1,1) dot (1,4,-1) = 9+4-1 ≠ 0
So the line intersects the plane at one point.