You can use a formula, although I think it's not too difficult to just go through the steps. I would draw a picture first:
You are given that $\vec{p} = (1,0,1)$ and you already found $\vec{m} = (1, -2, 4)$ and $\vec{l}_0 = (1,2,-1)$. Now it's a matter of writing an expression for $\vec{l}(t) - \vec{p}_0$:
\begin{align}
\vec{l}(t) - \vec{p}_0 =&\ (\ (t + 1) - 1\ ,\ (-2t + 2) - 0\ ,\ (4t - 1) - 1\ )\\
=&\ (\ t\ ,\ -2t + 2\ ,\ 4t - 2\ )
\end{align}
Now you dot this with the original slope of the line (recall that $\vec{l}(t) - \vec{p}_0$ is the slope of the line segment connecting the point and the line). When this dot product equals zero, you have found $t_0$ and thus $\vec{x}_0$:
\begin{align}
\vec{m} \circ (\vec{l}(t) - \vec{p}_0) =&\ (1,-2,4)\circ(\ t\ ,\ -2t + 2\ ,\ 4t - 2\ ) \\
=&\ t + 4t - 4 + 16t - 8 \\
=&\ 21t - 12
\end{align}
Setting this to $0$ gives that $21t_0 - 12 = 0 \rightarrow t_0 = \frac{4}{7}$. This gives the point $\vec{x}_0$ as:
\begin{align}
\vec{x}_0 =&\ \vec{l}(t_0) = (\ \frac{4}{7} + 1\ ,\ -\frac{8}{7} + 2\ ,\ \frac{16}{7} - 1\ ) \\
=&\ \frac{1}{7}(11, 6, 9)
\end{align}
So finally the distance would be the distance from $\vec{p}_0$ to $\vec{x}_0$:
\begin{align}
d =&\ \sqrt{\left(\frac{11}{7} - 1\right)^2 + \left(\frac{6}{7} - 0\right)^2 + \left(\frac{9}{7} - 1\right)^2}\\
=&\ \sqrt{\left(\frac{4}{7}\right)^2 + \left(\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2} \\
=&\ \frac{1}{7}\sqrt{4^2 + 6^2 + 2^2}\\
=&\ \frac{1}{7}\sqrt{56} \\
=&\ \frac{2}{7}\sqrt{14}
\end{align}
...or perhaps $\sqrt{\frac{8}{7}}$ is more appealing.
Extra Info
There's no need to worry about whether or not my 2D picture is really representative--it is. No matter how high the dimensions of the problem, the problem itself can always be mapped to exactly 2 dimensions unless the point is on the line--then it's a 1 dimensional problem--which of course we can represent in 2 dimensions just as we can represent this 2 dimensional problem in much higher ones.
Best Answer
The normal to the plane is ${\bf n}=(3,4,-1)$ as you have found. A vector in the direction of the line is ${\bf v}=(-2,3,-1)$. The angle between them is given by the dot product formula: $$\cos\theta=\frac{\bf n\cdot v}{|{\bf n}|\,|{\bf v}|}=\frac{7}{\sqrt{26}\sqrt{14}} =\frac{7}{2\sqrt{91}}=\frac{\sqrt{91}}{26}\ .$$ And the angle you want is $\frac\pi2-\theta$, draw a diagram and you will see why.