UPSC NDA I 2024 Mathematics Paper

Concept:The equation of a line through a point with given direction ratios can be written in parametric form using a parameter $t$.This parametric form is then substituted into the plane equation to find the value of $t$.Explanation:Let the line pass through $(1, -1, 2)$ with direction ratios $<3, 2, 2>$.Write the line in symmetric form and set it equal to $t$:$$\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{2} = t.$$From this, express $x$, $y$, $z$ in terms of $t$:$$x = 3t + 1,\quad y = 2t - 1,\quad z = 2t + 2.$$Let the point of intersection with the plane $x + 2y + 3z = 18$ be $P = (3t+1,\;2t-1,\;2t+2)$.Substitute into the plane equation:$$(3t+1) + 2(2t-1) + 3(2t+2) = 18.$$Simplify step by step:$$3t+1 + 4t-2 + 6t+6 = 18.$$Combine like terms:$$13t + 5 = 18.$$Solve for $t$:$$13t = 13 \quad\Rightarrow\quad t = 1.$$Now substitute $t=1$ back into the parametric coordinates:$$x = 3(1)+1 = 4,\quad y = 2(1)-1 = 1,\quad z = 2(1)+2 = 4.$$Thus the point of intersection is $(4, 1, 4)$.Answer:The point of intersection is $(4, 1, 4)$, which corresponds to Option C.