 Hello and welcome to the session. Let's discuss the following question. It says show that the plane to the points 1, 1, 1, 1, minus 1, 1 and minus 7, 3, minus 5 is perpendicular to the exact plane. So let's now move on to the solution. Now the equation of the plane passing through the point 1, 1, 1 is given by a into x minus 1 plus b into y minus 1 plus c into z minus 1 is equal to 0. Let us name this as 1. Now we are given that this plane passes through the points 1, minus 1, 1 and minus 7, minus 3, minus 5. Now 1 passes through the points 1, minus 1, 1 and minus 7, 3, minus 5. Therefore we have a into 1 minus 1 plus b into minus 1, minus 1 plus c into 1 minus 1 is equal to 0. So this implies minus 2b is equal to 0 and this implies b is equal to 0. Let us name this as 2. Now also the plane passes through the point minus 7, 3 minus 5. So we have a into minus 7 minus 1 plus b into 3 minus 1 plus c into minus 5 minus 1 is equal to 0. So we have minus 8a plus 2b minus 6c is equal to 0. Let us name this as 3. Now we will solve equation 2 and 3 for a, b, c by using the method of cross multiplication. So cross multiplying 2 and 3 we have a upon minus 1 into 0 minus 1 into 0 is equal to b upon 0 into 0 minus minus 3 into 0 is equal to c upon 1 into minus 3 minus minus 1 into 0. So this implies a upon 0 is equal to b upon 0 is equal to c upon minus 3. Let this be equal to k. So this implies a is equal to 0 into k that is 0, b is 0 and c is equal to minus 3k. Now putting these values of a, b, c in 1 to get the equation of the plane we have 0 into x minus 1 plus 0 into y minus 1 minus 3k into z minus 1 is equal to 0. So this equation becomes minus 3k into z minus 1 is equal to 0. So the equation of the plane is z minus 1 is equal to 0. Now we have to prove that the plane passing through these three given points is perpendicular to the x, z plane. Now the equation of x, z plane is y is equal to 0. Now we know that two planes a1x plus b1y plus c1z is equal to 0 and a2x plus b2y plus c2z is equal to 0 are perpendicular to each other if a1a2 plus b1b2 plus c1c2 is equal to 0. Now here we have two planes z minus 1 is equal to 0 and y is equal to 0 and we have to show that these two planes are perpendicular to each other. So here a1 is 0, b1 is 0, c1 is 1, a2 is 0, b2 is 1 that is here the coefficient of y and c2 is 0. So we have a1a2 plus b1b2 plus c1c2 is equal to 0 that is 0 plus 0 plus 0 is equal to 0. So this implies plane z minus 1 is equal to 0 is perpendicular to x, z plane. So this completes the question and the session. Bye for now. Take care. Have a good day.