The direction ratios (d.r.s.) of the normal to the plane passing through the origin and the line of intersection of the planes $x+2y+3z=4$ and $4x+3y+2z=1$ are

Suppose the line $\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-2z+\beta=0$. Then $(\alpha+\beta)$ is equal to ... .

If the lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2$ $(ab \neq 0)$ are coplanar,then

The equation of the plane passing through the intersection of the planes $x + 2y + z - 1 = 0$ and $2x + y + 3z - 2 = 0$ is perpendicular to the plane $x + y + z - 1 = 0$. If this plane is also parallel to the plane $x + ky + 3z - 1 = 0$,then the value of $k$ is:

If the distance of the point $(1, -2, 3)$ from the plane $x + 2y - 3z + 10 = 0$ measured parallel to the line $\frac{x-1}{3} = \frac{2-y}{m} = \frac{z+3}{1}$ is $\sqrt{\frac{7}{2}}$,then the value of $|m|$ is equal to ....... .

The vector equation of the plane passing through the intersection of the planes $\overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $\overrightarrow{r} \cdot (\hat{i} - 2\hat{j}) = -2$,and the point $(1, 0, 2)$ is: