2. Find the equation of the plane that passes through $(-3,-4,2),(-3,4,1)$, and $(1,1,-5)$. (6 pts)

Step 1 ：Given three points in the plane, p1 = (-3,-4,2), p2 = (-3,4,1), and p3 = (1,1,-5).

Step 2 ：Form two vectors that lie in the plane using these points. The first vector, v1, is formed by subtracting p1 from p2, giving v1 = (0,8,-1). The second vector, v2, is formed by subtracting p1 from p3, giving v2 = (4,5,-7).

Step 3 ：Find the normal vector to the plane by taking the cross product of v1 and v2. This gives the normal vector = (-51,-4,-32).

Step 4 ：The equation of a plane in 3D space is given by Ax + By + Cz = D, where A, B, and C are the coefficients of the normal vector to the plane, and D is a constant.

Step 5 ：Substitute the coordinates of any point in the plane (we can use p1) into the equation to solve for D. This gives D = -105.

Step 6 ：Substitute A, B, C, and D into the equation of the plane to get the final equation: \(-51x - 4y - 32z = -105\).

Step 7 ：Final Answer: The equation of the plane is \(\boxed{-51x - 4y - 32z = -105}\).