 Hello and welcome to the session. I am Deepika here. Let's discuss the question which says find the intercepts cut off by the plane 2x plus y minus z is equal to 5. Let us first understand the intercept form of the equation of a plane. Let the equation of the plane is ax plus by y plus cz plus d is equal to 0 where d is not equal to 0. Let the plane may intercepts a, b, c x, y and z axis respectively. Hence the plane needs x, y and z axis at a 0, 0, 0, b, 0 and 0, 0, c respectively. Hence the intercepts a, b, c can be written as a is equal to minus d upon a b is equal to minus d upon b and c is equal to minus d upon c. So this is a key idea behind that question. We will take the help of this key idea to solve the above question. So let's start the solution. Now the given equation of the plane is plus y minus z is equal to 5 or this can be written as 2x plus y minus z minus 5 is equal to 0. So this is of the form plus by plus cz plus d is equal to 0. Here we have a is equal to 2, b is equal to 1, c is equal to minus 1 and d is equal to minus 5. So according to our key idea the intercepts of by the plane are a is equal to minus d upon a which is equal to 5 over 2 and b is equal to minus d upon b which is equal to 5 and c is equal to minus d upon c which is minus 1. So this is equal to minus 5. Hence the intercepts cut off by the plane 2x plus 5 minus z is equal to 5 or 5 by 2 5 minus 5. So this is the answer for the above question. This completes our session. I hope the solution is clear to you. Bye and have a nice day.