The equation of the plane which is parallel to the plane $x - 2y + 2z = 5$ and whose distance from the point $(1, 2, 3)$ is $1$ is:

$A$ point lying on the plane that passes through the points $\hat{i}-\hat{j}+\hat{k}$,$\hat{i}-2\hat{j}+3\hat{k}$,and $\hat{i}+2\hat{j}-3\hat{k}$ is:

The vector equation of the plane which is at a distance of $ \frac{3}{\sqrt{14}} $ from the origin and the normal vector from the origin is $ 2 \hat{i}-3 \hat{j}+\hat{k} $ is:

$A$ plane passes through a fixed point $A(1, -2, 3)$. If the locus of the foot of the perpendicular to it from the point $B(0, 1, 2)$ is $x^2 + y^2 + z^2 + \alpha x + \beta y + \gamma z + \delta = 0$,then the value of $\alpha + \beta + \gamma + \delta$ is

In $3$-dimensional space,the equation $x^2-8x+12=0$ represents ....

The plane,passing through the points $(0,-1,2)$ and $(-1,2,1)$ and parallel to the line passing through $(5,1,-7)$ and $(1,-1,-1)$,also passes through the point.