Let P𝑃 be the plane passing through the points [1,−2,−1][1,−2,−1], [−2,3,−3][−2,3,−3], and [2,0,3][2,0,3].Where does the line [1,3,1]+s[2,−2,0][1,3,1]+𝑠[2,−2,0] intersect this plane?

To find the intersection of a line and a plane, we first need to find the equation of the plane.

Step 1: Find the vectors of the plane We can find two vectors in the plane by subtracting the coordinates of the given points. Let's call these vectors A and B.

A = [−2,3,−3] - [1,−2,−1] = [-3, 5, -2] B = [2,0,3] - [1,−2,−1] = [1, 2, 4]

Step 2: Find the normal vector of the plane The normal vector (N) to the plane is the cross product of A and B.

N = A x B = [-3, 5, -2] x [1, 2, 4] = [-28, -14, 7]

Step 3: Find the equation of the plane The equation of the plane can be found using the normal vector and a point on the plane. We can use the point [1,−2,−1] (let's call it P).

The equation of the plane is N . (r - P) = 0, where r = [x, y, z] is a general point on the plane.

So, -28(x - 1) - 14(y + 2) + 7(z + 1) = 0 Simplify this to get the equation of the plane: -28x - 14y + 7z = -42

Step 4: Find the intersection of the line and the plane The equation of the line is [1,3,1]+s[2,−2,0] = [1 + 2s, 3 - 2s, 1].

Substitute x = 1 + 2s, y = 3 - 2s, and z = 1 into the plane equation:

-28(1 + 2s) - 14(3 - 2s) + 7*1 = -42 Solve this equation to find the value of s.

Step 5: Substitute s into the line equation Once you have the value of s, substitute it back into the line equation to find the coordinates of the intersection point.