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

The plane $2x - y + z = 4$ intersects the line segment joining the points $A(a, -2, 4)$ and $B(2, b, -3)$ at the point $C$ in the ratio $2:1$. The distance of the point $C$ from the origin is $\sqrt{5}$. If $ab < 0$ and $P$ is the point $(a - b, b, 2b - a)$,then $CP^2$ is equal to:

The coordinates of the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane $2x - 3y - 6z = 49$ are

Find the equation of the plane passing through the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$.

If the line $\bar{r}=(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+\hat{\jmath}+2 \hat{k})$ is parallel to the plane $\bar{r} \cdot (3 \hat{\imath}-2 \hat{\jmath}+m \hat{k})=10$,then the value of $m$ is

If a plane $P$ passes through the points $(1,0,0)$ and $(0,1,0)$ and makes an angle $\frac{\pi}{4}$ with the plane $x+y=3$, then the direction ratios of a normal to that plane $P$ are