= Solution
Two directions within the plane are
$$
\mathbf f-\mathbf e=\left(\frac12,-\frac12,0\right),\qquad
\mathbf g-\mathbf e=\left(\frac12,0,-\frac12\right).
$$
Their <cross product> is $(1,1,1)/4$, so a <unit vector> normal to the plane is
$$
\boxed{\hat{\mathbf n}=\frac{(1,1,1)}{\sqrt3}}.
$$
Its negative is equally valid. An <equation of a plane through three points> in vector form is
$$
\boxed{\hat{\mathbf n}\cdot(\mathbf r-\mathbf e)=0},
$$
or, parametrically, $\mathbf r=\mathbf e+s(\mathbf f-\mathbf e)+t(\mathbf g-\mathbf e)$. Since the coordinate sum of $E$ is one, the <Cartesian equation of a plane> is
$$
\boxed{x+y+z=1}.
$$
The <distance from a point to a plane> is the absolute normal displacement, giving
$$
\boxed{d(O,EFG)=|\hat{\mathbf n}\cdot\mathbf e|=\frac1{\sqrt3}}.
$$
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