If the shortest distance between the lines $I_1$ and $I_2$ given by $\vec{r}=a\hat{i}+2\hat{j}-\hat{k}+\lambda (2\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=\hat{i}-\hat{j}+\hat{k}+\mu ( 2\hat{i}-\hat{j}+\hat{k})$ is $\sqrt{\frac{35}{6}}$ units, the values of 'a' can be :

0, -8

The correct answer is option (1) → 0, -8

$l_1:\vec{r}=a\hat{i}+2\hat{j}-\hat{k}+\lambda (2\hat{i}-\hat{j}+\hat{k})$

$l_2:\vec{r}=\hat{i}-\hat{j}+\hat{k}+\mu ( 2\hat{i}-\hat{j}+\hat{k})$

these are parallel lines

$\vec{a_1}=a\hat i+2\hat j-\hat k$

$\vec{a_2}=a\hat i+2\hat j-\hat k$

$\vec{a_2}-\vec{a_1}=(1-a)\hat i-3\hat j+2\hat k$

$\vec b=2\hat i-\hat j+\hat k$

shortest distance

$=\frac{|\vec b×(\vec{a_2}-\vec{a_1})|}{|\vec b|}⇒\vec b×(\vec{a_2}-\vec{a_1})=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&1\\1-a&-3&2\end{vmatrix}$

$⇒\frac{|-\hat i+(a+3)\hat j+(a+5)\hat k|}{\sqrt{6}}=\frac{\sqrt{35}}{\sqrt{6}}$

so squaring both sides

$(-1)^2+(a+3)^2+(a+5)^2=35$

$a^2+6a+a^2+10a+25+9=34$

$a^2+8a=0$

$a=0,-8$