Mathematics Question from JEE Main 2025 · 4 April, Shift 2

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