UPSC NDA Math Model Paper 4

We have to find the distance of the point (7, 14, 5) from the plane 2x+4y−z=2
Here, the equation of the plane can be re-written as 2x+4y−z−2=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=7,y1=14 and z1=5
So, the distance of the given point form the given plane =|

2×7+4×14−5−2

√22+(−1)2+(4)2

|
=|

63

√21

|
So, the distance between the plane and the point is 3√21
Hence, option A is the correct answer.