QJEEM24Ma4297: JEE Main 2024 Mathematics - Class 12 JEEM24Ma4297 Solution: If the shortest distance between the lines is , where , then…

$\begin{aligned}& L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k} \\& L_1=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+4 \hat{k}) \\& L_2: \vec{r}=2 \hat{i}+3 \hat{j}+5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k}) \\& \vec{a}_1=2 \hat{i}+\hat{j}+3 \hat{k} \\& \vec{a}_2=2 \hat{i}+3 \hat{j}+5 \hat{k} \\& \vec{a}_2-\vec{a}_1=2 \hat{j}+2 \hat{k} \\& \vec{b}_1=\hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}_2=2 \hat{i}+3 \hat{j}+\hat{k} \\& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & -3 & 4 \\2 & 3 & 1\end{array}\right|\end{aligned}$

$\begin{aligned}& \hat{i}(-3-12)-\hat{j}(1-8)+\hat{k}(3+6) \\& =-15 \hat{i}+7 \hat{j}+9 \hat{k} \\& \left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{225+49+81} \\& \left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\frac{14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \\& m+n=387\end{aligned}$