The distance between the planes $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$ is

The Cartesian equation of a plane parallel to the plane $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$ and at a distance of $2$ units from it is

The $x$-intercept of a plane $\pi$ passing through the point $(1,1,1)$ is $\frac{5}{2}$ and the perpendicular distance from the origin to the plane $\pi$ is $\frac{5}{7}$. If the $y$-intercept of the plane $\pi$ is negative and the $z$-intercept is positive,then its $y$-intercept is

If the foot of the perpendicular from $(0,0,0)$ to a plane is $(1,2,3)$, then the equation of the plane is

$A$ plane passes through $(1, -2, 1)$ and is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. The distance of the point $(1, 2, 2)$ from this plane is . . . . . . units.

Find the direction ratios of the normal to the plane passing through the points $(0, -1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{\pi}{4}$ with the plane $y - z + 5 = 0$.