Let the line L: x-1 2 = y-2 1 = z+1 2 intersect the plane 2x - y + z = d ( d > 0 ) at the point P . Let the point Q be the foot of the perpendicular from the point R(2, 3, 2) to the line L . If the area of the triangle PQR is 3 2 , then the value of d is equal to _____ .

Step-by-step solution

We are given the line L: x-1 2 = y-2 1 = z+1 2 = . Any point on L is S(2 +1, +2, 2 -1) . The direction ratios of L are 2, 1, 2 . The vector connecting R(2, 3, 2) to S is RS = 2 -1, -1, 2 -3 . Since Q is the foot of the perpendicular from R to L , RQ is perpendicular to L . Thus, the dot product is zero: 2(2 -1) + 1( -1) + 2(2 -3) = 0 4 - 2 + - 1 + 4 - 6 = 0 9 - 9 = 0 = 1 So, Q (3, 3, 1) . Now, we find the length RQ : RQ = (3-2)^2 + (3-3)^2 + (1-2)^2 = 1 + 0 + 1 = 2 . The area of the right-angled triangle PQR is giv