Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ is $\frac{1}{\sqrt{6}}$, be $\mathrm{a}, \mathrm{b},(\mathrm{a}<\mathrm{b})$. Then the length of the latus rectum of the ellipse $\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1$ is :

<p>$$\begin{aligned} & \frac{x+1}{3}=\frac{y}{4}=\frac{z}{5} ;(p \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\ & d=\left|\frac{(\vec{a}-\vec{b}) \cdot\left(\vec{p}_1 \times \vec{p}_2\right)}{\left|\vec{p}_1 \times \vec{p}_2\right|}\right|=\frac{1}{\sqrt{6}} \\ & \vec{a}-\vec{b}=(p+1) \hat{i}+2 \hat{j}+\hat{k} \\ & \vec{p}_1 \times \vec{p}_2=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{array}\right|=\hat{i}-2 \hat{j}+\hat{k} \\ & \frac{1}{\sqrt{6}}=\left|\frac{(p+1)-4+1}{\sqrt{6}}\right| \\ & =|p-2|=1 \Rightarrow p=3,1 \\ & a=1, b=3 \\ & \frac{x^2}{1}+\frac{y^2}{9}=1 \end{aligned}$$</p> <p>Length of $L R=\frac{2 a^2}{b}=\frac{2}{3}$</p>