Find an equation of the plane. the plane that passes through the point (3, 6, −1) and contains the line x = 4 − t, y = 2t − 1, z = −3t

To find the equation of the plane, we first need to find two direction vectors that are parallel to the plane.

1. We know that the plane contains the line x = 4 - t, y = 2t - 1, z = -3t. This line can be written in vector form as r(t) = <4 - t, 2t - 1, -3t>. The direction vector of this line, which is also parallel to the plane, is d1 = <-1, 2, -3>.

2. We also know that the plane passes through the point (3, 6, -1). We can find a vector from this point to any point on the line. Let's choose the point on the line when t = 0, which is (4, -1, 0). The vector from (3, 6, -1) to (4, -1, 0) is d2 = <4 - 3, -1 - 6, 0 - (-1)> = <1, -7, 1>.

3. Now we have two direction vectors d1 and d2 that are parallel to the plane. The normal vector of the plane is the cross product of these two vectors. Let's calculate it:

   n = d1 x d2 = <-1, 2, -3> x <1, -7, 1> = <(-3)(-7) - 21, (-1)1 - (-3)1, (-1)(-7) - 21> = <21 - 2, -1 + 3, 7 - 2> = <19, 2, 5>.

4. Now we have the normal vector of the plane and a point through which the plane passes. The equation of the plane can be written as n . (r - r0) = 0, where n is the normal vector, r is a general point on the plane, and r0 is a known point on the plane. Let's use the point (3, 6, -1) as r0. Then the equation of the plane is:

   <19, 2, 5> . <x - 3, y - 6, z - (-1)> = 0 19*(x - 3) + 2*(y - 6) + 5*(z + 1) = 0

So, the equation of the plane is 19x + 2y + 5z - 57 - 12 - 5 = 0, or 19x + 2y + 5z - 74 = 0.