Let M be the midpoint of the side PQ of a triangle PQR . The point M is the intersection of the lines x-1 2 = y 2 = z-2 1 and x-2 1 = y-3 -1 = z-4 -1 . If the centroid of PQR is G(1, 2, -1) , then the distance of the vertex R from the origin is

Step-by-step solution

Let the given lines be L₁: x-1 2 = y 2 = z-2 1 = k and L₂: x-2 1 = y-3 -1 = z-4 -1 = m . Any point on L₁ is (2k+1, 2k, k+2) and on L₂ is (m+2, -m+3, -m+4) . For the point of intersection M , we equate the coordinates: 2k+1 = m+2 2k - m = 1 2k = -m+3 2k + m = 3 Solving these two equations gives 4k = 4 k = 1 . Substituting k=1 into the coordinates of L₁ , we get M(3, 2, 3) . We know that the centroid G of a triangle divides the median from any vertex to the midpoint of the opposite side in the ratio 2:1 . Thus, G = R