Direction ratios of the line represented by the equations $x = ay + b$ and $z = cy + d$ are:

Line $L_1$ passes through the points $\hat{i}+\hat{j}$ and $\hat{k}-\hat{i}$. Line $L_2$ passes through the point $\hat{j}+2\hat{k}$ and is parallel to the vector $\hat{i}+\hat{j}+\hat{k}$. If $x\hat{i}+y\hat{j}+z\hat{k}$ is the point of intersection of the lines $L_1$ and $L_2$,then $(y-x)=$

The point having position vector $\vec{p} = 4\hat{i} - 11\hat{j} + 2\hat{k}$ lies on which of the following lines?

If the lines $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z + 1}{3}$ and $\frac{x + 2}{2} = \frac{y - k}{3} = \frac{z}{4}$ are coplanar,then the value of $k$ is

The point of intersection of the two lines $\frac{x - 3}{3} = \frac{y - 3}{-1}, z - 1 = 0$ and $\frac{x - 6}{2} = \frac{z - 1}{3}, y - 2 = 0$ is

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