g Let P be the plane that goes through the points A(1, 3, 2), B(2, 3, 0), and C(0, 5, 3). Let ` be the line through the point Q(1, 2, 0) and parallel to the line x = 5, y = 3−t, z = 6+2t. Find the (x, y, z) point of intersection of the line ` and the plane P.

Answer:   (x, y, z) = (1, 1/3, 3 1/3)Step-by-step explanation:The normal to plane ABC can be found as the cross product ...   AB×BC = (1, 0, 2)×(2, -2, -3) = (4, 1, 2)Then the equation of the plane is ...   4x +y +2z = 4·0 +5 +2·3 . . . . using point C to find the constant   4x +y +2z = 11__The direction vector of the reference line is the vector of coefficients of t: (0, -1, 2). Then the line through point Q is ...   (x, y, z) = (1, 2, 0) +t(0, -1, 2) = (1, 2-t, 2t)__The value of t that puts a point on this line in plane ABC can be found by substituting these values for x, y, and z in the plane's equation.   4(1) +(2 -t) +2(2t) = 11Solving for t gives ...   t = 5/3so the point of intersection of the plane and the line is   (x, y, z) = (1, 2-t, 2t) = (1, 2-5/3, 2·5/3) = (1, 1/3, 3 1/3)