The equation of plane equidistant from lines  $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $2x-y-4=0=3x-z-8$ is/are

Since, line  $L_1$  II  $L_2$. Hence, there will be infinite planes cutting line L and parallel to given line.
Equation of line  $L:\frac{x-\frac{1}{2}}{1}=\frac{y+1}{2}=\frac{z+\frac{5}{2}}{3}$
$\Rightarrow$ Plane  $P_1:2x-y-2=0$
Plane  $P_2:3x-z-4=0$
$\Rightarrow$ Required plane  $p:p_1+\lambda p_2=0$
$$$(2x-y-2)+\lambda (3x-z-4)=0,\lambda \in R$

Another is system of planes parallel to plane containing  $L_1$ and  $L_2$
Normal to lines  $L_1$ and  $L_2:\overrightarrow{x}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 6& 11\\ 1& 2& 3\end{array}\right|=-4\hat{i}+8\hat{j}-4\hat{k}$
⇒ One of D.R's normal  $=(1,-2,1)$
⇒ Equation of plane containing line's  $L_1$ and  $L_{2}x-2y+z=0$
⇒ Any plane parallel to the plane is  $x-2y+z=d,d\in R$