Final answer:
To find the equation of the plane, we use the cross product to find a normal vector to the plane from two direction vectors and then use point-normal form to determine the plane equation, resulting in 20x - 2y - 9z = 57.
Step-by-step explanation:
To find an equation of the plane that passes through the point (3, 6, -1) and contains the line x = 4 - t, y = 2t - 1, z = -3t, we need a point on the plane and a normal vector to the plane. We already have a point, (3, 6, -1). The direction vector of the line given can be taken as a direction vector of the plane. From the line equations, we can get the direction vector v = (-1, 2, -3).
To find another direction vector, we take a specific point on the line by plugging in a value for t. Let's use t = 0; this gives us the point (4, -1, 0). The vector from this point to (3, 6, -1) is u = (-1, 7, -1).
We now need a normal vector n which is perpendicular to both direction vectors u and v. This can be found using the cross product: n = u × v. Calculating the cross product yields n = (20, -2, -9).
Now, we can write the equation of the plane using the point-normal form. If (x, y, z) is any point on the plane and P0 = (3, 6, -1) is our given point, then n · ((x, y, z) - P0) = 0 gives the plane equation. Substituting and expanding, we get 20(x - 3) - 2(y - 6) - 9(z + 1) = 0, which simplifies to the plane equation 20x - 2y - 9z = 57.