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

To find the intersection of a line and a plane, we first need to find the equation of the plane. We can do this by finding the normal vector to the plane, which is the cross product of two vectors in the plane, and then using the point-normal form of the equation of a plane.

Step 1: Find two vectors in the plane We can get these by subtracting the coordinates of the given points: Vector A = [−2,1,−3] - [3,3,−3] = [-5, -2, 0] Vector B = [3,−3,2] - [3,3,−3] = [0, -6, 5]

Step 2: Find the normal vector to the plane This is the cross product of the two vectors we just found: Normal = A x B = [-25 - 0(-6), 00 - (-5)(-6), -5*(-6) - 0*0] = [-10, -30, 30]

Step 3: Find the equation of the plane We can use the point-normal form of the equation of a plane, which is (p - p0) . n = 0, where p is a point in the plane, p0 is a known point in the plane (we can use any of the given points), and n is the normal vector. Plugging in the values we have, we get: -10(x - 3) - 30(y - 3) + 30(z + 3) = 0 Simplifying, we get the equation of the plane: -10x - 30y + 30z = -90

Step 4: Find the intersection of the line and the plane The line is given by the parametric equations x = 1 + s, y = 1 - 2s, z = 1 - 2s. We can substitute these into the equation of the plane to find the value of s at the intersection point: -10(1 + s) - 30(1 - 2s) + 30(1 - 2s) = -90 Solving this equation for s, we get s = -1.

Step 5: Substitute s = -1 into the parametric equations of the line to find the intersection point: x = 1 + (-1) = 0 y = 1 - 2*(-1) = 3 z = 1 - 2*(-1) = 3

So, the line intersects the plane at the point (0, 3, 3).