VITEEE 2010 Paper

We have the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$...(i) and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$...(ii) Let a point (2r + 1, 3r - 1, 4r + 1) be on the line Eq. (i). If this is an intersection point of both the lines, then it will lie on Eq. (ii), also ∴ $\frac{2r+1-3}{1}=\frac{3r-1-k}{2}=\frac{4r+1}{1}$(iii) Taking first and third part of eq. (iii), we get $2r-2=4r+1$ ⇒ $r=-\frac{3}{2}$ Taking second and third part of eq. (iii), we get $3r-1-k=8r+2$ ⇒ $3r-1-k-8r-2=0$ ⇒ $k=-5r-3$ ⇒ $k=-5\left(-\frac{3}{2}\right)-3\;\;\;\;\;\left(\because r=-\frac{3}{2}\right)$ ⇒ $k=\frac{15}{2}-3$ ⇒ $k=\frac{9}{2}$