UPSC NDA I 2021 Mathematics Solved Paper

CONCEPT: Perpendicular distance of a plane $a x+b y+c z+d=0$ from a point $P(x_{1}, y_{1}, z_{1})$ is given by: $d=\left|\;\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$ CALCULATION: Let $A(x, y, z)$ be the foot of the perpendicular drawn fromthe origin to the plane $x+y+z=3$. As we know that, the perpendicular distance of a plane $a x+b y+c z+d=0$ from a point $P(x_{1}, y_{1}, z_{1})$ is given by: $d=\left|\;\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2+b^{2}+c^{2}}}}\right|$ So, the distance betweenthe origin and the plane $x+y+z-3=0$ is given by: $d=\left|\;\frac{0+0+0-3}{\sqrt{1^{2}+1^{2}+1^{2}}}\right|=\;\frac{3}{\sqrt{3}}=\sqrt{3}$ So, this means that the length of the line joining the points origin and $A$ is $\sqrt{3}$ As we can see that from the given options, if $A=(1,1,1)$then the distance between the points origin and $A$ is $\sqrt{3}$ Hence, correct option is 3 .