Let the shortest distance between the lines $L:\frac{x-5}{-2}=\frac{y-\lambda }{0}=\frac{z+\lambda }{1},\lambda \ge 0$ and $L_1:x+1=y-1=4-z\text{ be }2\sqrt{6}\text{. If }(\alpha ,\beta ,\gamma )$lies on $\mathrm{L}$, then which of the following is NOT possible?

$$\begin{gathered} L:\frac{x-5}{-2}=\frac{y-\lambda }{0}=\frac{z+\lambda }{1};\lambda \ge 0 \\ {L}_1:\frac{x+1}{1}=\frac{y-1}{1}=\frac{z-4}{-1} \end{gathered}$$
Shortest distance $=\frac{\left|\left|\begin{array}{ccc}5+1& \lambda -1& -\lambda -4\\ -2& 0& 1\\ 1& 1& -1\end{array}\right|\right|\begin{array}{} \\ \\ \end{array}}{\sqrt{(0-1{)}^2+(1-2{)}^2+(-2{)}^2}}=2\sqrt{6}$
$\Rightarrow |\lambda -3|=12\Rightarrow \lambda =15,-9;\Rightarrow \lambda =15 (\because \lambda \ge 0)$
$\therefore \mathrm{L}\text{ is }\frac{x-5}{-2}=\frac{y-15}{0}=\frac{z+15}{1}$

Any point on L is of the form
$(5-2t,15,-15+t)\Rightarrow \alpha +2\gamma \ne 24$
All other options are possible for some t