The largest value of $a$, for which the perpendicular distance of the plane containing the lines $\vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+a \hat{j}-\hat{k})$ and $\vec{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-a \hat{k})$ from the point $(2,1,4)$ is $\sqrt{3}$, is _________.

<p>Normal to plane $$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & a & { - 1} \cr { - 1} & 1 & { - a} \cr } } \right|$$</p> <p>$= \widehat i(1 - {a^2}) - \widehat j( - a - 1) + \widehat k(1 + a)$</p> <p>$= (1 - a)\widehat i + \widehat j + \widehat k$</p> <p>$\therefore$ Plane $(1 - a)(x - 1) + (y - 1) + z = 0$</p> <p>Distance from (2, 1, 4) is $\sqrt 3$ i.e.</p> <p>$$ \Rightarrow \left| {{{(1 - a) + 0 + 4} \over {\sqrt {{{(1 - a)}^2} + 1 + 1} }}} \right| = \sqrt 3 $$</p> <p>$\Rightarrow 25 + {a^2} - 10a = 3{a^2} - 6a + 9$</p> <p>$\Rightarrow 2{a^2} + 4a - 16 = 0$</p> <p>$\Rightarrow {a^2} + 2a - 8 = 0$</p> <p>$a = 2$ or $- 4$</p> <p>$\therefore$ ${a_{\max }} = 2$</p>