Find the equation of a plane that is at a unit distance from the origin and is parallel to the plane $x - 2y + 2z - 5 = 0$.

The distance of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{m} = q$ is given by $\frac{|\vec{a} \cdot \vec{m} - q|}{|\vec{m}|}$. If the distance of the point $\hat{i} + 2\hat{j} + 3\hat{k}$ from the plane $\vec{r} \cdot (2\hat{i} + 6\hat{j} - 9\hat{k}) = -1$ is $p$ and the distance of the origin from this plane is $q$,then $p - q =$

For scalars $\lambda, \mu$, if the vector equation of a plane is $r=(2+3 \lambda-\mu) \hat{i}+(1-2 \lambda+3 \mu) \hat{j}+(-2+2 \lambda+\mu) \hat{k}$, then its Cartesian equation is

The equation of the plane passing through $(-1, 1, 2)$,whose normal makes equal acute angles with the coordinate axes is

The plane passing through the points $(1,1,1)$,$(1,-1,1)$ and $(-7,-3,-5)$ is

The equation of the plane passing through the points $(2,3,1)$ and $(4,-5,3)$ and parallel to the $y$-axis is: