 Hi, and welcome to the session. Here it says, class the following question. The question says, if a and b be the points 3, 4, 5, and minus 1, 3, minus 7 respectively, find the equation of the set of points b, such that b a squared plus b b squared is equal to k squared, where k is a constant. Let's now begin with this illusion. In this question, we are given two points a and b. And we are asked to find the equation of the set of points b, such that b a squared plus b b squared is equal to a squared. Let b, having coordinates x, y, z, be the required point. By the distance formula, b a squared is equal to x minus 3 whole square plus y minus 4 whole square plus z minus 5 whole square. And b b squared is equal to x plus 1 whole square plus y minus 3 whole square plus z plus 7 whole square. We are given that b a squared plus b b squared is equal to k squared. So this means x minus 3 whole square plus y minus 4 whole square plus z minus 5 whole square plus x plus 1 whole square plus y minus 3 whole square plus z plus 7 whole square is equal to k squared. Now this implies x squared minus 6x plus 9 plus y squared minus a2 plus 16 plus z squared minus 10z plus 25 plus x squared plus 2x plus 1 plus y squared minus 6y plus 9 plus z squared plus 14z plus 49 is equal to k squared. This implies 2x squared plus 2y squared plus 2z squared minus 4x minus 14y plus 4z plus 109 is equal to k squared. This implies 2 into x squared plus y squared plus 6 squared minus 2x minus 7y plus 2z is equal to k squared minus 109. This implies x squared plus y squared plus z squared minus 2x minus 7y plus 2z is equal to k squared minus 109 by 2. Hence the required equation is x squared plus y squared plus x squared minus 2x minus 7y plus 2z equals to k squared minus 109 by 2. This is our required answer. So this completes the session. I end here.