= Solution
Use only the <dot product> to minimize the squared <Euclidean distance> from the origin:
$$
D(t)^2=(\boldsymbol a_2+t\boldsymbol n_2)\cdot(\boldsymbol a_2+t\boldsymbol n_2)
=\|\boldsymbol a_2\|^2+2t(\boldsymbol a_2\cdot\boldsymbol n_2)+t^2\|\boldsymbol n_2\|^2.
$$
Completing the square gives
$$
D(t)^2=\|\boldsymbol n_2\|^2
\left(t+\frac{\boldsymbol a_2\cdot\boldsymbol n_2}{\|\boldsymbol n_2\|^2}\right)^2
+\|\boldsymbol a_2\|^2-
\frac{(\boldsymbol a_2\cdot\boldsymbol n_2)^2}{\|\boldsymbol n_2\|^2}.
$$
The square is minimized at $t_0=-(\boldsymbol a_2\cdot\boldsymbol n_2)/\|\boldsymbol n_2\|^2$, yielding
$$
\boxed{d_0=\sqrt{\boldsymbol a_2\cdot\boldsymbol a_2
-\frac{(\boldsymbol a_2\cdot\boldsymbol n_2)^2}{\boldsymbol n_2\cdot\boldsymbol n_2}}.}
$$
Geometrically this subtracts the squared length of the <orthogonal projection> parallel to the line. At the minimizing point, $(\boldsymbol a_2+t_0\boldsymbol n_2)\cdot\boldsymbol n_2=0$, as required for perpendicular distance.
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