The Cartesian equation of the line which passes through the points $(3,1,2)$ and $(-1,2,1)$ is

$A$ line drawn from a point $A(-2,-2,3)$ and parallel to the line $\frac{x}{-2}=\frac{y}{2}=\frac{z}{-1}$ meets the $YOZ-$ plane in point $P$. Then,the coordinates of the point $P$ are:

If the foot of the perpendicular drawn from the point $A(1, 0, 3)$ on a line passing through $B(\alpha, 7, 1)$ is $P\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$,then $\alpha$ is equal to:

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - c}{2} = \frac{z}{1}$ intersect, then the radius of the circle $x^2 + y^2 - 4x + 10y + c = 0$ is

Let $D$ be the foot of the perpendicular drawn from the point $A(2,0,3)$ to the line joining the points $B(0,4,1)$ and $C(-2,0,4)$. Then the ratio in which $D$ divides $BC$ is

Let the values of $p$,for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{r}=(p\hat{i}+2\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$ is $\frac{1}{\sqrt{6}}$,be $a$ and $b$ $(a < b)$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is: