Equation of a Line in Space

The vector equation of the line passing through the point $(-1,5,4)$ and perpendicular to the plane $z=0$ is 

The equations of the straight line passing through $P(\alpha , \beta , \gamma )$ and parallel to the $x$-axis are

The vector equation of the line through the points $(3, 4, -7)$ and $(1, -1, 6)$ is

The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is 

The vector equation of a line passing through the point $\left(-1,-1,2\right)$ and parallel to the line $2x-2=3y+1=6z-2$ is

The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

The equation of the line passing through the points $a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ is

The direction ratios of the line $x-y+z-5=0=x-3y-6$ are proportional to

The shortest distance between the lines $\overrightarrow{r}=\left(6\hat{i}+5\hat{j}+6\hat{k}\right)+\lambda \left(5\hat{i}-16\hat{j}+7\hat{k}\right)$ and $\overrightarrow{r}=\left(9\hat{i}+13\hat{j}+15\hat{k}\right)+\mu \left(3\hat{i}+8\hat{j}-5\hat{k}\right)$ is

The vector form of the line  $\frac{x-3}{-2}=\frac{y-4}{-5}=\frac{x+7}{13}$ is 

The foot of the perpendicular from the point $A(7,14,5)$ to the plane $2x+4y-z=2$ is

The straight line $\frac{x-2}{3}=\frac{y-3}{1}=\frac{z+1}{0}$ is

The vector equation of the $x-axis$ is given by

The coordinates of the point where the line through the points $A(5,1,6)$and $B(3,4,1)$ crosses the $yz$-plane is

A line passes through the point $A(-2,4,-5)$ and is parallel to the line $\frac{x+3}{-3}=\frac{y-4}{5}=\frac{z+8}{6}$. The vector equation of the line is

A line passes through the points $A(2,-1,4)$ and $B(1,2,-2) .$ The equations of the line $AB$ are

The equation of the line passing through the points $a_1\widehat{\mathbf{i}}+a_2\widehat{\mathbf{j}}+a_3\widehat{\mathbf{k}} \mathrm{a}\mathrm{n}\mathrm{d} b_1\widehat{\mathbf{i}}+b_2\widehat{\mathbf{j}}+b_3\widehat{\mathbf{k}}$ is