UPSC NDA I 2021 Mathematics Solved Paper

Concept:Three points in 3D are collinear if the vectors joining them are proportional.Explanation:Let the points be $A(k,1,3)$, $B(1,-2,k+1)$, and $C(15,2,-4)$.The direction ratios of $AB$ are $(1-k,\ -3,\ k-2)$.The direction ratios of $AC$ are $(15-k,\ 1,\ -7)$.For $C$ to lie on line $AB$, these ratios must be proportional:$\frac{15-k}{1-k} = \frac{1}{-3} = \frac{-7}{k-2}$.Equating the first two fractions: $\frac{15-k}{1-k} = \frac{1}{-3}$ Cross-multiplying: $-3(15-k) = 1-k \ \Rightarrow\ -45+3k = 1-k \ \Rightarrow\ 4k = 46 \ \Rightarrow\ k = \frac{23}{2}$.Equating the last two fractions: $\frac{1}{-3} = \frac{-7}{k-2}$ Cross-multiplying: $k-2 = 21 \ \Rightarrow\ k = 23$.Since $k$ cannot be both $\frac{23}{2}$ and $23$ at the same time, no single value of $k$ satisfies both equalities.Thus, there is no real value of $k$ for which the line passes through the given point.Answer:Zero (Option A).