The square of the distance of the point $\mathrm{P}(5,6,7)$ from the line $\frac{x-2}{2}=\frac{y-5}{3}=\frac{z-2}{4}$ is equal to:

$\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(a \hat{i}-\hat{j}), a \neq 0$ and $\vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+a \hat{k})$ from the origin is :

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)