Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

<p>$$\begin{aligned} & L_1: \frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1} \\ & L_2: \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4} \end{aligned}$$</p> <p>$$\begin{aligned} & a_1:(3, \alpha, 3) \quad a^2=(-3,-7, \beta) \\ & \vec{b}_1=3 \hat{i}-\hat{j}+\hat{k} \quad \vec{b}_2=-3 \hat{i}+2 \hat{j}+4 \hat{k} \end{aligned}$$</p> <p>$d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|}=3 \sqrt{30}$</p> <p>$=\frac{\left|\begin{array}{ccc}6 & \alpha+7 & 3-\beta \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|}{\sqrt{6^2+15^2+3^2}}=3 \sqrt{30}$</p> <p>$$\begin{aligned} \Rightarrow & |-15 \alpha-3 \beta-132|=270 \\ & |5 \alpha+\beta+44|=90 \\ \Rightarrow & 5 \alpha+\beta=90-44=46 \end{aligned}$$</p>