Let $\mathrm{H}: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is $\frac{8}{3}$. If the line $x=\alpha$ intersects the hyperbola H at the points A and B such that the area of the triangle AOB is $4 \sqrt{15}$, where O is the origin, then $\alpha^2$ equals

The shortest distance between the lines

$$ \vec{r}=\left(\frac{1}{3} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\frac{8}{3} \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) $$

and $\vec{r}=\left(-\frac{2}{3} \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{k}}\right)+\mu(\hat{\mathrm{j}}-\hat{\mathrm{k}}), \lambda, \mu \in \mathbb{R}$, is:

If $\left(2 \alpha+1, \alpha^2-3 \alpha, \frac{\alpha-1}{2}\right)$ is the image of $(\alpha, 2 \alpha, 1)$ in the line $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z}{1}$, then the possible value(s) of $\alpha$ is (are)

Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u} \times \hat{v}|=\frac{\sqrt{3}}{2}$. If $\overrightarrow{\mathrm{A}}=\lambda \hat{u}+\hat{v}+(\hat{u} \times \hat{v})$, then $\lambda$ is equal to: