 Hello and welcome to the session. In this session we will discuss a question which says that find the equation of the plane through the points 3, 2, minus 2 and minus 2, 4, 5 and is perpendicular to the plane 2x minus 3y plus 5z is equal to 10. Now before starting the solution of this question we should know some results. First is the equation of a plane passing through the points x1, y1, z1 is a into x minus x1 the whole plus b into y minus y1 the whole plus c into z minus z1 the whole is equal to 0. Let this be equation number 1 and if 1 passes through x2, y2, z2 then the equation will be a into x2 minus x1 the whole plus b into y2 minus y1 the whole plus c into z2 minus z1 the whole is equal to 0. And secondly the planes a1x plus b1y plus c1z plus d1 is equal to 0 and a2x plus d2y plus c2z plus d2 is equal to 0 are perpendicular a1 into a2 plus b1 into b2 plus c1 into c2 is equal to 0. Now these results will work out as a key idea for solving out this question and now we will start with the solution. Now we have to find the equation of the plane through these points and which is perpendicular to this plane. Now given the plane is passing through the point 3 2 minus 2 so using this result which is given in the key idea the equation of any plane through the point 3 2 minus 2 is given by a into x minus 3 the whole plus b into z minus of minus 2 which will be c into z plus 2 the whole is equal to 0. Now let this be equation number one since it passes through the point minus 2 4 5 therefore the equation of the plane now by using this result which is given in the key idea now this equation of the plane which is given by equation number one passes through this point that is minus 2 4 5 therefore the equation of the plane will be a into minus 2 minus 3 the whole plus b into 4 minus 2 the whole plus c into 5 plus 2 the whole is equal to 0 this implies minus 5 in plus 2 b plus 7 c is equal to 0. Now let this be equation number two. Now this is the equation number one which is the equation of the plane now it is given let the plane which is given by equation number one is perpendicular to the plane 2 x minus 3 y plus 5 z is equal to 10. Now using the condition of perpendicularity since the plane which is given by equation number one is perpendicular to this plane therefore a 1 a 2 that is a into 2 plus b 1 b 2 that is plus b into minus 3 plus c 1 c 2 that is plus c into 5 is equal to 0. Now this implies 2 a minus 3 b plus 5 c is equal to 0 that is the equation number three. So this is the equation number two and this is the equation number three from two and three we have minus 5 a plus 2 b plus 7 c is equal to 0 and 2 a minus 3 b plus 5 c is equal to 0. Now solving these two equations by the method of cross multiplication we get a over 5 into 2 is 10 minus 7 into minus 3 is minus 21 is equal to b over 7 into 2 is 14 minus minus 5 into minus 5 is minus 25 is equal to c over 15 minus 2 into 2 is 4. Now this implies a over 10 plus 21 is 31 is equal to b over 14 plus 25 is 39 is equal to c over minus 4 is equal to 11. Now let this be equal to a constant say lambda. So this implies a is equal to 31 lambda b is equal to 39 lambda and c is equal to 11 lambda. Now this is the equation number one. Now putting the values of a b and c in equation number one we get 31 lambda into x minus 3 the whole plus 39 lambda into y minus 2 the whole plus 11 lambda into z plus 2 the whole is equal to 0. Now dividing throughout by lambda this implies 31 into x minus 3 the whole plus 39 into y minus 2 the whole plus 11 into z plus 2 the whole is equal to 0 which further implies 31 x minus 93 plus 39 y minus 78 plus 11 z plus 22 is equal to 0 which implies 31 x plus 39 y plus 11 z is equal to 149. So this is the required equation of the plane and this is the solution of the given question and that's all for this so if you want have enjoyed the session.