If the shortest distance between the lines (x-lambda / 3)=(y-2 / -1)=(z-1 / 1) and (x+2 / -3)=(y+5 / 2)=(z-4 / 4) is frac44sqrt(30), then the largest possible value of |lambda| is equal to _________.

L_1: (x-lambda / 3)=(y-2 / -1)=(z-1 / 1) \ L_2: (x+2 / -3)=(y+5 / 2)=(z-4 / 4) \ n_1 x n_2=|\ccc hat i hat j hat k \ 3 -1 1 \ -3 2 4 \| \ =-6 hat i-15 hat j+3 hat k \ d=|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||=frac44sqrt(30) \ |frac-6 lambda-12-105-9sqrt(270)|=frac44sqrt(30) \ |6 lambda+126|=132 \ |lambda+21|=22 \ lambda+21= ± 22 \ |lambda|_max =43 \

If the shortest distance between the lines frac xlambda... | null (Mathematics)

\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}

\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}

Explanation

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