 Hi friends, I am Purva and today we will discuss the following question. Find the equation of the plane which is perpendicular to the plane 5x plus 3y plus 6z plus 8 is equal to 0 and which contains the line of intersection of the planes x plus 2y plus 3z minus 4 is equal to 0 and 2x plus y minus z plus 5 is equal to 0. Let us begin with the solution now. Now the vector equation of a plane that passes through the intersection of two given planes a1x plus b1y plus c1z plus d1 is equal to 0 and a2x plus b2y plus c2z plus d2 is equal to 0 is given by a1x plus b1y plus c1z plus d1 plus lambda into a2x plus b2y plus c2z plus d2 is equal to 0. So here the equation of the plane containing the line of intersection of the planes x plus 2y plus 3z minus 4 is equal to 0 and 2x plus y minus z plus 5 is equal to 0 is Now applying this formula we get x plus 2y plus 3z minus 4 plus lambda times 2x plus y minus z plus 5 is equal to 0 or we can write this as Now taking out x common from these both terms we get x into 1 plus 2 lambda plus taking out y common from these both terms we get y into 2 plus lambda plus now taking out z common from these both terms we get z into 3 minus lambda minus 4 plus 5 into lambda is equal to 0 We mark this as equation 1 Now we have to find the equation of the plane which is perpendicular to the plane 5x plus 3y plus 6z plus 8 is equal to 0 and which contains the line of intersection of the planes x plus 2y plus 3z minus 4 is equal to 0 and 2x plus y minus z plus 5 is equal to 0 Now the equation of the plane containing the line of intersection of these two planes is this and we also have that this plane is perpendicular to the plane 5x plus 3y plus 6z plus 8 is equal to 0 So now we know that this plane is perpendicular to the plane 5x plus 3y plus 6z plus 8 is equal to 0 Now we know that if the two planes a1x plus b1y plus c1z plus d1 is equal to 0 and a2x plus b2y plus c2z plus d2 is equal to 0 are perpendicular to each other then we have a1 into a2 plus b1 into b2 plus c1 into c2 is equal to 0 Now we have this plane 1 is perpendicular to this plane So applying this formula we get 5 into 1 plus 2 lambda plus 3 into 2 plus lambda plus 6 into 3 minus lambda is equal to 0 Now here we have a1 is equal to 1 plus 2 lambda b1 is equal to 2 plus lambda c1 is equal to 3 minus lambda and d1 is equal to minus 4 plus 5 lambda and here a2 is equal to 5 b2 is equal to 3 c2 is equal to 6 and d2 is equal to 8 So putting all these values in this formula we get 5 into 1 plus 2 lambda plus 3 into 2 plus lambda plus 6 into 3 minus lambda is equal to 0 or we can write this as now opening the brackets we get 1 into 5 is 5 plus 5 into 2 lambda is 10 lambda plus 3 into 2 is 6 plus 3 into lambda is 3 lambda plus 6 into 3 is 18 6 into minus lambda is minus 6 lambda is equal to 0 or we have now solving this we get 29 plus 7 lambda is equal to 0 or we can write this as 7 lambda is equal to minus 29 therefore we have lambda is equal to minus 29 upon 7 Now putting lambda is equal to minus 29 upon 7 in this equation 1 we get x into 1 plus 2 lambda now lambda is equal to minus 29 upon 7 plus y into 2 plus lambda that is 2 minus 29 upon 7 plus z into 3 minus lambda now lambda is minus 29 upon 7 so here we get 3 plus 29 upon 7 plus minus 4 plus 5 lambda that is 5 into minus 29 upon 7 is equal to 0 or we have x into 1 minus 58 upon 7 plus y into 2 minus 29 upon 7 plus z into 3 plus 29 upon 7 minus 4 minus 145 upon 7 is equal to 0 or we have 51 x plus 15 y minus 50 z plus 173 is equal to 0 solving this equation we get 51 x plus 15 y minus 50 z plus 173 is equal to 0 therefore the required equation of the plane is 51 x plus 15 y minus 50 z plus 173 is equal to 0 this is our answer hope you have understood the solution bye and take care