JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Three Dimensional Geometry

First, find the equation of the plane containing the two given lines. Both lines pass through points $(1,2,3)$ and $(2,3,4)$ respectively, with direction vectors $\overrightarrow{d_1}=(2,3,4)$ and $\overrightarrow{d_2}=(3,4,5)$.

The normal to the plane is $\vec{n}=\overrightarrow{d_1}\times \overrightarrow{d_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)=(-1,2,-1)$.

The plane equation is $-1(x-1)+2(y-2)-1(z-3)=0$, simplifying to $-x+2y-z=0$ or $x-2y+z=0$.

The given plane is $Ax-2y+z=d$. For these to be parallel, $A=1$. The distance between parallel planes $x-2y+z=0$ and $x-2y+z=d$ is $\frac{|d|}{\sqrt{1+4+1}}=\frac{|d|}{\sqrt{6}}=\sqrt{6}$.

Thus $|d|=6$.