UPSC NDA I 2021 Mathematics Solved Paper

Concept:The foot of the perpendicular from a point to a plane is found by moving along the plane's normal direction until the plane equation is satisfied.Explanation:The given plane is $x + y + z = 3$.The normal vector to the plane is $(1, 1, 1)$.Any line through the origin along this normal can be written as $(x, y, z) = (t, t, t)$ for some parameter $t$.This line meets the plane where the point satisfies $x+y+z=3$.Substitute $(t, t, t)$: $t + t + t = 3$.So $3t = 3$, hence $t = 1$.Thus the foot of the perpendicular is $(1, 1, 1)$.Check: Distance from origin to $(1,1,1)$ is $\sqrt{1^2+1^2+1^2} = \sqrt{3}$, which equals the perpendicular distance from origin to plane $\left|\frac{-3}{\sqrt{3}}\right|=\sqrt{3}$.Among the options, only $(1,1,1)$ satisfies the plane equation and lies on the normal line.Answer:C. $(1, 1, 1)$