Final answer:
To find the equation of the plane that passes through the given point and contains the given line, we find the normal vector by taking the cross product of two non-parallel vectors on the plane. Then, we use the point-normal form of the equation to find the equation of the plane. The equation is -3x - 3y - 2z + 22 = 0.
Step-by-step explanation:
To find an equation of the plane that passes through the point (3,5,-1) and contains the line x=4 - t, y=2t - 1, z=-3t, we need to determine the normal vector of the plane. The direction ratios of the line are the coefficients of t in the x, y, and z equations. So, the direction ratios of the line are -1, 2, and -3. To find the normal vector, we take any two non-parallel vectors on the plane and find their cross product. Let's take the vectors (4, 2, -3) and (1, 0, 0) as our two vectors. The cross product of these vectors is (0, -3, -2). Therefore, the normal vector of the plane is (0, -3, -2).
Now that we have a point on the plane (3,5,-1) and the normal vector (0, -3, -2), we can use the point-normal form of the equation of a plane to find the equation. The equation is given by:
0(x - 3) - 3(y - 5) - 2(z + 1) = 0
Simplifying the equation, we get:
-3x + 9 - 3y + 15 - 2z - 2 = 0
-3x - 3y - 2z + 22 = 0
So, the equation of the plane that passes through the point (3,5,-1) and contains the line x=4 - t, y=2t - 1, z=-3t is -3x - 3y - 2z + 22 = 0.