Let $L$ be the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=1$. If $L$ makes an angle $\alpha$ with the positive $x$-axis,then $\cos \alpha$ equals:

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^o$ about its line of intersection with the plane $5x - 4y - 2z + 1 = 0$. The equation of the plane in its new position is:

The distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is:

If the plane passing through the points $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}-\hat{k}$ and the origin meets the line passing through the points $\hat{i}+3\hat{j}-2\hat{k}$ and $\hat{i}-\hat{j}+3\hat{k}$ at the point $A$, then $A=$

The angle between the line $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3\hat{i}+\hat{j})$ and the plane $\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k})=8$ is:

The equation of the plane which is parallel to the line $\frac{x - 4}{1} = \frac{y + 3}{-4} = \frac{z + 1}{7}$ and passes through the points $(0, 0, 0)$ and $(3, -1, 2)$ is