Given, $L_1:\frac{x-1}{\lambda }=\frac{y-2}{1}=\frac{z-3}{2}$

and $L_2:\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda }$ are coplanar

Now using the condition of coplanarity we get,

$\Rightarrow \left|\begin{array}{ccc}27& 20& 31\\ \lambda & 1& 2\\ -2& 3& \lambda \end{array}\right|=0\Rightarrow \lambda =3$

Now, normal of plane $P$, which contains $L_1$ and $L_2$ is given by

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ -2& 3& 3\end{array}\right|=-3\hat{i}-13\hat{j}+11\hat{k}$

Now equation of required plane $P$ will be

$-3\left(x-1\right)-13\left(y-2\right)+11\left(z-3\right)=0$

 $\Rightarrow 3x+13y-11z+4=0$

Now satisfy all given points from option we get, $\left(0,4,5\right)$ does not lie on plane $P$.