If the line x-3 2 = y+5 -1 = z+2 2 lies in the plane x+3 y-z+ =0 , then values of and respectively are ....

Step-by-step solution

The line L with symmetric equation x-3 2 = y+5 -1 = z+2 2 contains point P(3, -5, -2) with direction vector d = 2, -1, 2 . The plane : x + 3y - z + = 0 has normal vector n = , 3, -1 . For L to lie in , the direction vector must be perpendicular to the plane's normal vector: d n = 2 + (-1)(3) + 2(-1) = 2 - 5 = 0 Solving yields = 5 2 . Additionally, point P must satisfy the plane equation: 3 + 3(-5) - (-2) + = 3 - 13 + = 0 Substituting = 5 2 gives 15 2 - 13 + = 0 , hence = 11 2 . The values = 5 2 and = 11 2 correspon