UPSC NDA Math Model Paper 9

We have to find the distance of the point (2, 1, 1) from the plane x - 2y + 4z + 8 = 0
As we know that, the distance between the point and the plane is given by |

Ax1+By1+Cz1−d

√A2+B2+C2

|
Here, x1=2,y1=1 and z1=1
So, the distance of the given point form the given plane = |

1×2−2×1+4×1+8

√12+(−2)2+(4)2

|
=|

12

√21

|
So, the distance between the plane and the point is

4√21

7