$$\begin{aligned} & \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \\ & \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5} \end{aligned}$$

the shortest distance between the lines

$$=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right)}{\left|\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right|}\right|$$

$$=\left|\frac{\left|\begin{array}{ccc}\lambda-4 & 0 & 2 \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}\right|$$

$$=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right|$$

$$\begin{aligned} & \frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right| \\ & 3=|\lambda-4| \\ & \lambda-4= \pm 3 \\ & \lambda=7,1 \end{aligned}$$

Sum of all possible values of $$\lambda$$ is $$=8$$