 Good morning friends. I am Poorva and today we will discuss the following question. Find the equation of the plane through the line of intersection of the planes x plus y plus z is equal to 1 and 2x plus 3y plus 4z is equal to 5 which is perpendicular to the plane x minus y plus z is equal to 0. Now the equation of the 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 and c2z 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. Also if the 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. So this is the key idea behind our question. Let us begin with the solution now. Now the two planes are x plus y plus z is equal to 1 and 2x plus 3y plus 4z is equal to 5 or we can write this as x plus y plus z minus 1 is equal to 0 and 2x plus 3y plus 4z minus 5 is equal to 0. Now by k-idea we know that the equation of the plane that passes through the line of intersection of two given planes is given by Now by k-idea the equation of the plane that passes through the intersection of these two planes is given by this equation. So, here we get the equation of the plane that passes through the intersection of these two planes is given by x plus y plus z minus 1 plus lambda into 2x plus 3y plus 4z minus 5 is equal to 0. Or we can write this equation as x into 1 plus 2 lambda plus y plus 2 y into 1 plus 3 lambda plus z into 1 plus 4 lambda minus 1 plus 5 lambda is equal to 0. We mark this as equation 1. Now this equation 1 is an equation of the plane as it is of the form a1x plus b1y plus c1z plus d1 is equal to 0. Now in the question we are given that this equation of the plane is perpendicular to the plane x minus y plus z is equal to 0. So as equation 1 is perpendicular to the plane x minus y plus z is equal to 0, therefore by key idea we know that if the two planes are perpendicular then we have a1 a2 plus b1 c1 c2 is equal to 0. So here we get 1 plus 2 lambda into 1 plus 1 plus 3 lambda into minus 1 plus 1 plus 4 lambda into 1 is equal to 0. This implies 1 plus 2 lambda minus 1 minus 3 lambda plus 1 plus 4 lambda is equal to 0 and this implies 3 lambda plus 1 is equal to 0 which implies lambda is equal to minus 1 upon 3. Now putting lambda is equal to minus 1 upon 3 in equation 1 we get x into 1 plus 2 into minus 1 upon 3 plus y into 1 plus 3 into minus 1 upon 3 plus z into 1 plus 4 into minus 1 upon 3 minus 1 plus 5 into minus 1 upon 3 is equal to 0. Solving this equation we get this implies x into 1 upon 3 plus y into 0 plus z into minus 1 upon 3 minus minus 2 upon 3 is equal to 0 and this implies x minus z plus 2 is equal to 0. Thus we have got the equation of the plane as x minus z plus 2 is equal to 0. This is our answer. Hope you have understood the solution. Bye and take care.