 Hi friends, I am Purva and today we will work out the following question. In the following, find the distance of the given point from the corresponding given plane. The point is 2, 3 minus 5 and the plane is x plus 2y minus 2z is equal to 9. Now, let the equation of the plane be ax plus by plus cz is equal to d and the given point be x1, y1, z1. Then the distance of this point from the plane is given by mod of ax1 plus by1 plus cz1 minus d upon under root of a square plus b square plus c square. So, this is the key idea behind our question. Let us begin with the solution now. Now, we are given the equation of the plane as x plus 2y minus 2z is equal to 9 and the given point is 2, 3 minus 5. Now, comparing this equation of the plane with ax plus by plus cz is equal to d, we see that here a is equal to 1, b is equal to 2, c is equal to minus 2 and d is equal to 9. Also, the point x1, y1, z1 is 2, 3 minus 5. Now, by key idea we know that the distance of the point x1, y1, z1 from the plane ax plus by plus cz is equal to d is given by mod of ax1 plus by1 plus cz1 minus d upon under root of a square plus b square plus c square. So, here we get required distance is equal to mod of ax1 plus by1 plus cz1 minus d upon under root of a square plus b square plus c square. This is equal to mod of ax1 is equal to 1 into 2 plus by1 is equal to 2 into 3 plus cz1 is equal to minus 2 into minus 5 minus d that is minus 9 upon under root of a square that is 1 square plus b square that is 2 square plus c square that is minus 2 whole square. This is equal to mod of 1 into 2 is 2 plus 2 into 3 is 6 plus minus 2 into minus 5 is plus 10 minus 9 upon under root of 1 plus 4 plus 4. This is equal to mod of 9 upon under root 9 and this is equal to 9 upon under root 9 which is equal to 9 upon 3 and this is equal to 3. So, we have got the required distance as 3. This is our answer. Hope you have understood the solution. Bye and take care.