Let the line x-2 3 = y-1 -5 = z+2 2 lie in the plane x+3 y- z+ =0 , then the value of ( - ) is equal to

Step-by-step solution

Determining - for a line lying in a plane The line x-2 3 = y-1 -5 = z+2 2 has direction vector d = 3, -5, 2 and passes through point P₀(2, 1, -2) . The plane x + 3y - z + = 0 has normal vector n = 1, 3, - . For the line to lie in the plane, the direction vector must be orthogonal to the normal vector: d n = 0 . 3 1 + (-5) 3 + 2 (- ) = 0 3 - 15 - 2 = 0 -2 = 12 = -6 Additionally, point P₀ must satisfy the plane equation: 2 + 3(1) - (-6)(-2) + = 0 2 + 3 - 12 + = 0 -7 + = 0 = 7 The value - = 7 - (-6) = 13 . Final answe