Solution
= Solution
The normal from (ii) gives the <plane> equation $x+y+2z=3$. The <distance from a point to a plane> is the absolute value of the normalized plane equation at that point. At the origin this gives
$$
\boxed{d=\frac3{\sqrt{1^2+1^2+2^2}}=\frac3{\sqrt6}=\frac{\sqrt6}{2}}.
$$
For a geometric check, the perpendicular foot is $H=(1/2,1/2,1)$, parallel to the normal and satisfying the plane equation. Its <Euclidean norm> is the same distance.