Let the vertex A of a triangle ABC be the point of intersection of the lines x-1 1 = y-2 2 = z-3 3 and x-1 -1 = y-2 1 = z-3 1 . If the midpoint of the side BC is M(7, 8, 0) , then the square of the perpendicular distance of the centroid of ABC from the plane containing the two given lines is

Step-by-step solution

From the equations of the given lines, it is clear that they both pass through the point A(1, 2, 3) . Thus, the vertex A is (1, 2, 3) . The direction ratios of the two lines are d₁ = i + 2 j + 3 k and d₂ = - i + j + k . The normal vector to the plane containing these lines is: n = d₁ d₂ = vmatrix i & j & k 1 & 2 & 3 -1 & 1 & 1 vmatrix = i (2-3) - j (1+3) + k (1+2) = - i - 4 j + 3 k The equation of the plane passing through A(1, 2, 3) with normal n is: -1(x - 1) - 4(y - 2) + 3(z - 3) = 0 -x + 1 - 4y + 8 + 3z - 9 = 0