 Hello and welcome to the session. In this session we discuss the following question which says, find the equation of the plane passing through the point minus 1, minus 1, 2 and perpendicular to each of the following planes 2x plus 3y minus 3z equal to 2 and 5x minus 4y plus z equal to 6. Before moving on to the solution, let's recall the equation of the plane passing through the point, say a point p which coordinates x1, y1, z1 is given by a into x minus x1 plus b into y minus y1 plus c into z minus z1 equal to 0 where we have this a, b, c are the direction ratios of the normal to the plane. If we consider the equation of two planes as a1x plus b1y plus c1z plus d1 equal to 0 and a2x plus b2y plus c to z plus d2 equal to 0 where the direction ratios of the normal to the planes are a1, b1, c1 and a2, b2, c2 respectively. Then we say if these two planes are perpendicular to each other then we have a1, a2 plus b1, b2 plus c1, c2 is equal to 0. This is the key idea to be used for this question. Now let's proceed with the solution. First we find the equation of the plane passing through say a point p which coordinates minus 1, minus 1, 2. Now this would be given by a into x plus 1 plus b into y plus 1 plus c into z minus 2 equal to 0. We have done this using the key idea. Now this could also be written as ax plus by plus cz equal to minus a minus b plus 2c. Let this be equation 1. Now here we have a, b and c are the direction ratios of the normal to the plane 1. Now the other equations of the planes given to us are 2x plus 3y minus 3z equal to 2. Let this be equation 2 then 5x minus 4y plus z equal to 6. Let this be equation 3. Now it is given in the question that we have that the plane 1 is perpendicular to the plane 2. So now using the key idea and these two equations of the plane we have the condition 2a plus 3v minus 3c equal to 0. Let this be equation 4. Then we also have that plane 1 is perpendicular to the plane 3. So using the key idea and the two equations that is plane 1 and plane 3 we get the condition that is 5a minus 4b plus c equal to 0. Let this be equation 5. Now we need to solve equation 4 and equation 5 so as to get the values for a, b and c. So now solving equations 4 and 5 by cross multiplication we get a upon 3 into 1 3 minus minus 4 into minus 3 that is 12 equal to b upon 5 into minus 3 that is minus 15 minus 2 into 1 that is 2 equal to c upon 2 into minus 4 that is minus 8 minus 5 into 3 that is 15. So this gives us a upon minus 9 equal to b upon minus 17 equal to c upon minus 23 that is we can also say a upon 9 equal to b upon 17 equal to c upon 23 let this be equal to say lambda. So this gives us a is equal to 9 lambda b is equal to 17 lambda and c is equal to 23 lambda. Now substituting the values of a, b and c in 1 we get 9 lambda x plus 17 lambda y plus 23 lambda z equal to minus 9 lambda minus 17 lambda plus 46 lambda. Now taking lambda common from both the sides and lambda lambda gets cancelled so this would give us 9 x plus 17 y plus 23 z equal to minus 9 minus 17 plus 46. Thus we get 9 x plus 17 y plus 23 z equal to 20 or we can say 9 x plus 17 y plus 23 z minus 20 equal to 0 is the required equation of the plane. Thus the final answer is required equation of the plane is 9 x plus 17 y plus 23 z minus 20 equal to 0. So this completes the session. Hope you have understood the solution for this question.