 Hi and welcome to the session. Let's work out the following question. The question says find the equation of the plane passing through the point 111 and perpendicular to each of the following planes that is x plus 2y plus 3z is equal to 7 and 2x minus 3y plus 4z equals to 0. Let us start with the solution to this question. We see that the equation of the plane passing through the point 111 is a into x minus 1 plus b into y minus 1 plus c into z minus 1 is equal to 0. We call this the first equation as this plane is perpendicular to x plus 2y plus 3z equals to 7. Therefore, a plus 2b plus 3c is equal to 0. This we call 2. Similarly, this plane is perpendicular to 2x minus 3y plus 4z equal to 0. Therefore, 2a minus 3b plus 4c is equal to 0 and this we call 3. Now solving 2 and 3 we get a divided by 8 plus 9 because 4 2s are 8 minus minus 9 is plus 9. This we are doing by cross multiplication is equal to b divided by 1 into 4 is 4 and 3 into 2 is 6. So 6 minus 4 is equal to c divided by minus 3 minus 4 is equal to lambda c. This implies a divided by 17 is equal to b divided by 2 is equal to c divided by minus 7. So this implies that a is equal to 17 lambda, b is equal to 2 lambda and c is equal to minus 7 lambda. Now putting these values in the first equation we get 17 lambda into x minus 1 plus 2 lambda into y minus 1 minus 7 lambda into z minus 1 is equal to 0. This implies lambda into 17x minus 17 plus 2y minus 2 minus 7z plus 7 is equal to 0. Since lambda is not equal to 0, here we have taken lambda to be some number that is not equal to 0. Therefore 17x plus 2y minus 7z is equal to 12. So the required plane is 17x plus 2y minus 7z is equal to 12 and this is also our answer to this question. I hope that you understood the solution and enjoyed the session. Have a good day.