The shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\mathrm{d}=\left|\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|}{\sqrt{\left(\mathrm{a}_1 \mathrm{~b}_2-\mathrm{a}_2 \mathrm{~b}_1\right)^2+\left(\mathrm{a}_1 \mathrm{c}_2-\mathrm{a}_2 \mathrm{c}_1\right)^2+\left(\mathrm{b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right)^2}}\right|$
$\therefore \quad d=\left|\frac{\left|\begin{array}{lll}1 & 2 & 2 \\ 2 & 3 & \lambda \\ 1 & 4 & 5\end{array}\right|}{\sqrt{(8-3)^2+(10-\lambda)^2+(15-4 \lambda)^2}}\right|$
$$
\begin{aligned}
& \Rightarrow \frac{1}{\sqrt{3}}=\left|\frac{5-2 \lambda}{\sqrt{17 \lambda^2-140 \lambda+350}}\right| \\
& \Rightarrow \frac{1}{3}=\frac{25-20 \lambda+4 \lambda^2}{17 \lambda^2-140 \lambda+350} \\
& \Rightarrow 5 \lambda^2-80 \lambda+275=0 \\
& \Rightarrow \lambda^2-16 \lambda+55=0 \\
& \Rightarrow(\lambda-5)(\lambda-11)=0 \\
& \Rightarrow \lambda=5 \text { or } \lambda=11
\end{aligned}
$$
$\therefore \quad$ Sum of possible values of $\lambda=16$