UPSC NDA Math Model Paper 1

Here, we have to find the distance of the point (2, 3, 4) from the plane 3x - 6y + 2z + 11 = 0
As we know that, the perpendicular distance of a plane ax + by + cz + d = 0 from a point P (x1,y1,z1) is given by: D =|

ax1+by1+cz1+d

√a2+b2+c2

|
Here, x1=2,y1=3,z1=4,a=3,b=−6 and c=2So, the required distance between the given point and plane is D=|

3×2−6×3+2×4+11

√32+(−6)2+22

|
⇒D=1
Hence, option A is the correct answer.