Let the vertex A of an isosceles triangle ABC ( AB = AC ) be (8, 1, 1) . The base BC lies on the line x-2 2 = y+1 3 = z-4 6 . If the area of ABC is 49 sq. units, then the sum of the squares of the coordinates of the vertices B and C is _______.

Step-by-step solution

Let the given line be L: x-2 2 = y+1 3 = z-4 6 = . Any point on L is of the form P(2 +2, 3 -1, 6 +4) . If P is the foot of the perpendicular from A(8, 1, 1) to L , then the direction ratios of AP are (2 -6, 3 -2, 6 +3) . Since AP is perpendicular to L , we have: 2(2 -6) + 3(3 -2) + 6(6 +3) = 0 4 - 12 + 9 - 6 + 36 + 18 = 0 49 = 0 = 0 . So, the foot of the perpendicular is P(2, -1, 4) . The height of the triangle is AP = (2-8)^2 + (-1-1)^2 + (4-1)^2 = 36 + 4 + 9 = 7 . Given the area of ABC is 49 : 1 2 BC AP = 49 1 2