UPSC NDA I 2024 Mathematics Paper

Concept:The perpendicular distance from the origin to a plane is found using the distance formula after determining the equation of the plane.Explanation:The plane passes through the three given points: $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.These are the intercept points on the $x$-, $y$-, and $z$-axes respectively.Therefore, the equation of the plane in intercept form is $\frac{x}{1} + \frac{y}{1} + \frac{z}{1} = 1$.Multiplying through gives $x + y + z = 1$, or equivalently $x + y + z - 1 = 0$.Let $p$ be the perpendicular distance from the origin $(0,0,0)$ to this plane.Using the distance formula: $p = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$, where $a=1,b=1,c=1,d=-1$.Substitute: $p = \frac{|0 + 0 + 0 - 1|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}}$.Then $3p^2 = 3 \times \left(\frac{1}{\sqrt{3}}\right)^2 = 3 \times \frac{1}{3} = 1$.Answer:$3p^2 = 1$, which corresponds to option D.