If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to _________.

<p>$$\begin{aligned} & L_1: \frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1} \\ & L_2: \frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4} \end{aligned}$$</p> <p>$$\begin{aligned} & n_1 \times n_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{array}\right| \\ & =-6 \hat{i}-15 \hat{j}+3 \hat{k} \\ & d=\left|\frac{[(\lambda+2) \hat{i}+7 \hat{j}-3 \hat{k}][-6 \hat{i}-15 \hat{j}+3 \hat{k}]}{|-6 \hat{i}-15 \hat{j}+3 \hat{k}|}\right|=\frac{44}{\sqrt{30}} \\ & \left|\frac{-6 \lambda-12-105-9}{\sqrt{270}}\right|=\frac{44}{\sqrt{30}} \\ & |6 \lambda+126|=132 \\ & |\lambda+21|=22 \\ & \lambda+21= \pm 22 \\ & |\lambda|_{\max }=43 \end{aligned}$$</p>