The shortest distance from the origin to the plane $2x+3y+6z-21=0$ is

Given, equation of the plane is,

 $2x+3y+6z-21=0$ 

The formula to find the distance from a point can be calculated as,

$d=\left|\frac{A{x}_1+B{y}_1+C{z}_1-D}{\sqrt{A^2+B^2+C^2}}\right|$

Thus, the shortest distance of the given place form the origin (0, 0, 0) is,

$\begin{array}{l}\left|\frac{2\left(0\right)-3\left(0\right)+6\left(0\right)+21}{\sqrt{2^2+(3{)}^2+6^2}}\right|\\ =\frac{21}{\sqrt{4+9+36}}\\ =\frac{21}{\sqrt{49}}\\ =\frac{21}{7}=3\end{array}$