Let the shortest distance between the lines

$L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$ and

$L_{1}: x+1=y-1=4-z$ be $2 \sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$,

then which of the following is NOT possible?

$\frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$ <br/><br/>$$ \begin{aligned} & \frac{x+1}{1}=\frac{y-1}{1}=\frac{z-4}{-1} \\\\ & \vec{a}_{1}=5 \hat{i}+\lambda \hat{j}-\lambda \hat{k}_{,} \vec{a}_{2}=-\hat{i}+\hat{j}+4 \hat{k} \\\\ & \vec{a}_{1}-\vec{a}_{2}=6 \hat{i}+(\lambda-1) \hat{j}-(\lambda+4) \hat{k} \\\\ & \vec{b}_{1}=-2 \hat{i}+\hat{k}, \vec{b}=\hat{i}+\hat{j}-\hat{k} \\\\ & \vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{array}\right| \\ & =-\hat{i}-\hat{j}-2 \hat{k} \end{aligned} $$ <br/><br/>$\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot \vec{b}_{1} \times \vec{b}_{2}=-6+1-\lambda+2 \lambda+8=\lambda+3$ <br/><br/>and $\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{6}$ <br/><br/>$\because \frac{|\lambda+3|}{\sqrt{6}}=2 \sqrt{6}$ <br/><br/>$\therefore \lambda=9, \because \lambda \geq 0$ <br/><br/>$\therefore \quad L: \frac{x-5}{-2}=\frac{y-9}{0}=\frac{z+9}{1}=k$ <br/><br/>$\therefore \quad \alpha=-2 k+5, \beta=9, \gamma=k-9$ <br/><br/>Here $k$ is real then <br/><br/>$\alpha+2 \gamma=-13 \neq 24$. <br/><br/>But all other are in terms of $k$ hence possible.