KEAM 2013 Math Paper

Given lines are, $\;\frac{x+1}{2} = \;\frac{-y+1}{3} = \;\frac{z+1}{-2} = (r_{1}) \;\;$ (say).....(i) So, $(2 r_{1}-1, 1-3 r_{1}, -2 r_{1}-1)$ be any point on line (1) and $\;\; \;\frac{x-3}{1} = \;\frac{y-\lambda}{2} = \;\frac{z}{3} = r_{2} \;\;$ (say) ......(ii) So, $(r_{2}+3, 2 r_{2}+\lambda, 3 r_{2})$ be any point on line (ii). since, both lines intersect each other. $\therefore 2 r_{1}-1 = r_{2}+3$ $\Rightarrow \;\; 2 r_{1}-r_{2}=4$.......(iii) $-3 r_{1}+1 = 2 r_{2}+\lambda$ $\Rightarrow \;\; -3 r_{1}-2 r_{2}=\lambda-1$.....(iv) and $\;\; -2 r_{1}-1 = 3 r_{2}$ $\Rightarrow \;\; 2 r_{1}+3 r_{2}=-1$.....(v) On subtracting Eq. (v) from Eq. (iii), we get $-4 r_{2}=5 \Rightarrow r_{2}=-\frac{5}{4}$ From Eq. (iii), we get $2 r_{1}=4-\;\frac{5}{4} = \;\frac{11}{4}$ $\Rightarrow \;\; r_{1}=\;\frac{11}{8}$ On putting the value of $r_{1}$ and $r_{2}$ in Eq. (iv), we get $-3\left(\;\frac{11}{8}\right)-2\left(-\;\frac{5}{4}\right)=\lambda-1$ $\Rightarrow \;\; \lambda-1=-\;\frac{33}{8}+\;\frac{20}{8}=\;\frac{-13}{8}$ $\therefore \;\; \lambda=1-\;\frac{13}{8}=\;\frac{-5}{8}$