Let $A$ and $B$ be two points on the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{-1}$. If the distance of point $P(1, 1, 1)$ from the points $A$ and $B$ is $\sqrt{3}$,then the distance between $A$ and $B$ is:

Let a line passing through the point $P(4,1,0)$ intersect the line $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{array}\right|$ is equal to

Find the shortest distance between the lines whose vector equations are $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$.

The length of the perpendicular from the point $(1, -2, 5)$ to the line passing through $(1, 2, 4)$ and parallel to the line $x + y - z = 0 = x - 2y + 3z - 5$ is:

The lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular to each other. Then,the value of $p$ is . . . . . . .

If $(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2})$ 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)