If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is

<p>$$\begin{aligned} & d=\left|\frac{(\vec{a}-\vec{b}) \cdot \vec{p}_1 \times \vec{p}_2}{\left|\vec{p}_1 \times \vec{p}_2\right|}\right|=\frac{5}{\sqrt{6}} \\ & \vec{a}-\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \\ & \overrightarrow{p_1} \times \overrightarrow{p_2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & \alpha & 1 \end{array}\right| \\ & \quad=\hat{i}(3-4 \alpha)-\hat{j}(-2)+\hat{k}(2 \alpha-3) \\ & (\vec{a}-\vec{b}) \cdot\left|\overrightarrow{p_1} \times \overrightarrow{p_2}\right|=3-4 \alpha+4-4 \alpha+6 \\ & =13-8 \alpha \end{aligned}$$</p> <p>$$\begin{aligned} & \left|\frac{13-8 \alpha}{\sqrt{(3-4 \alpha)^2+4+(2 \alpha-3)^2}}\right|=\frac{5}{\sqrt{6}} \\ & \left|\frac{13-8 \alpha}{\sqrt{20 \alpha^2-36 \alpha+22}}\right|=\frac{5}{\sqrt{6}} \\ & =6(13-8 \alpha)^2=25\left(20 \alpha^2-36 \alpha+22\right) \\ & =116 \alpha^2+348 \alpha-464=0 \\ & \text { Sum of roots }=-3 \end{aligned}$$</p>