Let the shortest distance between the lines  $L:\frac{x-5}{-2}=\frac{y-\lambda }{0}=\frac{z+\lambda }{1},\lambda \ge 0$   and  $L_1:x+1=y-1=4-z\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}be3\sqrt{6}If(\alpha ,\beta ,\gamma )$  lies on L then which of the following is Not possible?

$$\begin{gathered} r_1=(5,\lambda ,-\lambda )+t(-2,0,1)=\overline{a}+t\overline{b} \\ {r}_2=(-1,1,-4)+s(1,1,-1)=\overline{c}+s\overline{d} \\ \overline{b}\times \overline{d}=-i-J-2k \\ \overline{a}-\overline{c}=-6i+(1-\lambda )J+(4+\lambda )\overline{k} \\ S.D=\left|\frac{(\overline{a}-\overline{c}),(\overline{b}\times \overline{d})}{|\overline{b}\times \overline{d}|}\right|=3\sqrt{6} \\ \lambda =15 \\ \therefore \alpha =-2t+5,\beta =15,\gamma =t-15 \\ \alpha +2\gamma =-25 \\ 2\alpha +\gamma =-4t+10+t-15=-3t-5 \\ =-5t+25 \end{gathered}$$