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(B) The two lines pass through the point $P(1, 1, 0)$. The direction vectors of the lines are $\vec{v}_1 = \hat{i} + a\hat{j} - \hat{k}$ and $\vec{v}_

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$2$

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(B) The two lines pass through the point $P(1, 1, 0)$. The direction vectors of the lines are $\vec{v}_1 = \hat{i} + a\hat{j} - \hat{k}$ and $\vec{v}_2 = -\hat{i} + \hat{j} - a\hat{k}$.The normal vector $\vec{n}$ to the plane containing these lines is given by the cross product $\vec{v}_1 	imes \vec{v}_2$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & a & -1 \ -1 & 1 & -a \end{vmatrix} = \hat{i}(-a^2 + 1) - \hat{j}(-a - 1) + \hat{k}(1 + a) = (1-a^2)\hat{i} + (a+1)\hat{j} + (1+a)\hat{k}$.Dividing by $(1+a)$ (assuming $a 
eq -1$),we get the normal vector $\vec{n}' = (1-a)\hat{i} + \hat{j} + \hat{k}$.The equation of the plane passing through $(1, 1, 0)$ with normal $\vec{n}'$ is:$(1-a)(x-1) + 1(y-1) + 1(z-0) = 0 \implies (1-a)x + y + z + a - 2 = 0$.The perpendicular distance from $(2, 1, 4)$ to the plane is given as $\sqrt{3}$:$\frac{|(1-a)(2) + 1 + 4 + a - 2|}{\sqrt{(1-a)^2 + 1^2 + 1^2}} = \sqrt{3}$.$\frac{|2 - 2a + 5 + a - 2|}{\sqrt{a^2 - 2a + 1 + 2}} = \sqrt{3} \implies \frac{|5 - a|}{\sqrt{a^2 - 2a + 3}} = \sqrt{3}$.Squaring both sides: $(5-a)^2 = 3(a^2 - 2a + 3)$.$25 - 10a + a^2 = 3a^2 - 6a + 9$.$2a^2 + 4a - 16 = 0 \implies a^2 + 2a - 8 = 0$.$(a+4)(a-2) = 0$,so $a = 2$ or $a = -4$.The largest value of $a$ is $2$.

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...

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