The square of the distance of the point P(5, 6, 7) from the line x-2 2 = y-5 3 = z-2 4 is equal to:

Step-by-step solution

Let the foot of the perpendicular from P(5, 6, 7) to the given line be Q . Any point on the line x-2 2 = y-5 3 = z-2 4 = is Q(2 + 2, 3 + 5, 4 + 2) . The direction ratios of PQ are (2 - 3, 3 - 1, 4 - 5) . Since PQ is perpendicular to the line, the dot product of their direction ratios is zero: 2(2 - 3) + 3(3 - 1) + 4(4 - 5) = 0 4 - 6 + 9 - 3 + 16 - 20 = 0 29 - 29 = 0 = 1 Substituting = 1 , the coordinates of Q are (4, 8, 6) . The square of the distance PQ^2 is (5 - 4)^2 + (6 - 8)^2 + (7 - 6)^2 = 1 + 4 + 1 = 6 . Answ