JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Three Dimensional Geometry

The plane containing the two given lines must pass through a point common to them, so first fix $\lambda$ by forcing the lines to meet. Writing Line 1 as $(2s-1,4s+3,3s-1)$ and Line 2 as $(2t-3,6t-2,\lambda t+1)$ and equating the first two coordinates gives $s=-\frac{1}{2}$, $t=\frac{1}{2}$, and the point of intersection $(-3,-2,1)$. This plane is parallel to $23x-10y-2z+48=0$, so the distance between them is just the distance of $(-3,-2,1)$ from that plane: $\frac{|23(-3)-10(-2)-2(1)+48|}{\sqrt{23^2+10^2+2^2}}=\frac{|-3|}{\sqrt{633}}=\frac{3}{\sqrt{633}}$. Comparing with $\frac{k}{\sqrt{633}}$ gives $k=3$.