Given $\bar{a}=a_1 i+a_2 \mathrm{j}+a_3 k$ and $\bar{b}=\sqrt{2} i+3 \sqrt{2} i+4 k$
Where, $\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3$ and $|\mathrm{a}|$ are rational numbers.
Now acc to question
$\begin{aligned} & \cot \left(\frac{A}{2}\right) \cdot \cot \left(\frac{c}{2}\right)=\frac{2 a b+a^2+b^2-c^2}{2 b c-b^2-c^2+a^2} \\ & =\frac{(2 b) a+\left(a^2-c^2\right)+b^2}{(2 b) c-b^2-c^2+a^2} \\ & =\frac{(a+c) a+a^2-c^2+\frac{(a+c)^2}{4}}{(a+c) c-\frac{(a+c)^2}{4}-c^2+a^2}\end{aligned}$
$\Rightarrow 2 \mathrm{a}_1+6 \mathrm{a}_2+4 \sqrt{3} \mathrm{a}_3=|\mathrm{a}| .6$
Since $a_1 a_2, a_3,|a|$ are rational numbers thus both sides should be rational $\Rightarrow a_3=0$ hence a lies in XY place.
Now $\cot \left(\frac{\mathrm{A}}{2}\right) \cdot \cot \left(\frac{\mathrm{c}}{2}\right)=\frac{2 \mathrm{ab}+\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2}{2 \mathrm{bc}-\mathrm{b}^2-\mathrm{c}^2+\mathrm{a}^2}$
$\begin{aligned} & =\frac{(2 b) a+\left(a^2-c^2\right)+b^2}{(2 b) c-b^2-c^2+a^2} \\ & =\frac{(a+c) a+a^2-c^2+\frac{(a+c)^2}{4}}{(a+c) c-\frac{(a+c)^2}{4}-c^2+a^2}\end{aligned}$
$\begin{aligned} & =\frac{4\left(a^2+a c+a^2-c^2\right)+a^2+c^2+2 a c}{4\left(a c+c^2\right)-\left(a^2+c^2+2 a c\right)-4 c^2+4 a^2} \\ & =\frac{4 a^2+4 a c+4 a^2-4 c^2+a^2+c^2+2 a c}{4 a c+4 c^2-a^2-c^2-2 a c-4 c^2+4 a^2} \\ & =\frac{9 a^2-3 c^2+6 a c}{3 a^2-c^2+2 a c} \\ & =\frac{3\left(3 a^2-c^2+2 a c\right)}{\left(3 a^2-c^2 4+2 a c\right)}=3\end{aligned}$