Step 1: Key Formula or Approach:
For a line through a point $A$ with direction vector $\vec{b}$, the distance of a point $P$ from the line is:
\[ d = \frac{|\overrightarrow{AP} \times \vec{b}|}{|\vec{b}|} \]
Step 2: Pick the data:
The line has denominators $0, 1, 0$, so it passes through $A(0,0,0)$ with $\vec{b} = \hat{j}$.
Then $\overrightarrow{AP} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Step 3: Find the cross product:
\[ \overrightarrow{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 1 & 0 \end{vmatrix} = -3\hat{i} + 0\hat{j} + 1\hat{k} \]
Its magnitude is $\sqrt{9 + 0 + 1} = \sqrt{10}$.
Step 4: Divide by the magnitude of the direction vector:
Here $|\vec{b}| = 1$, so $d = \sqrt{10}$.
Check: the line is the y-axis, so the distance is $\sqrt{x^2 + z^2} = \sqrt{1 + 9} = \sqrt{10}$.
Step 5: Check the options:
$2$ and $6$ do not match the value $\sqrt{10} \approx 3.16$.
$\sqrt{14}$ is the distance from the origin to $P$, which is larger than the perpendicular distance.
Final Answer:
The length of the perpendicular is $\sqrt{10}$, option (C).
\[ \boxed{\sqrt{10}} \]