I think you have the general idea, although it's not quite correct. Take any point on the line, say ##\mathbf{r}_0 = (0,9,0)##, as you chose. The moment of ##\mathbf{F}## about this point is
\begin{align*}
\mathbf{G} = (\mathbf{r} - \mathbf{r}_0) \times \mathbf{F} = (6, -9, -1) \times (2,0,6) = (-54, -38, 18)
\end{align*}The unit vector along the line, as you wrote, is ##\mathbf{n} = \frac{1}{\sqrt{58}} (7,0,-3)##. The moment of ##\mathbf{F}## about the line is the projection of ##\mathbf{G}## onto ##\mathbf{n}##, which is a scalar:
\begin{align*}
\mathbf{G} \cdot \mathbf{n} = \frac{1}{\sqrt{58}} (-54, -38, 18) \cdot (7,0,-3) = \dots
\end{align*}(It doesn't matter which point on the line you choose, because if you consider the point, say, ##\mathbf{r}_0 + \alpha \mathbf{n}##, then you have an extra term ##-\alpha \mathbf{n} \times \mathbf{F}## in the moment which vanishes upon taking the scalar product with ##\mathbf{n}##.)