Final answer:
To find the equation of the plane, we first find two points on the line and then find the normal vector by taking the cross product. The equation of the plane is -2x - y -11z + 37 = 0.
Step-by-step explanation:
To find an equation of the plane that passes through (3, 6, -1) and contains the line x = 4 - t, y = 2t - 1, z = -3t, we can use the point-normal form of the equation for a plane.
First, let's find two points on the line. When t = 0, we get the point (4, -1, 0). When t = 1, we get the point (3, 1, -3). These two points are on the plane we are trying to find.
Next, we can find the normal vector of the plane by taking the cross product of the vectors formed by subtracting the two points from the given point (3, 6, -1). Using the cross product, we find that the normal vector is (-2, -1, -11).
Finally, we can write the equation of the plane as -2(x - 3) - 1(y - 6) - 11(z + 1) = 0, or the simplified form: -2x - y -11z + 37 = 0.