In general, we know that the equation of the line passing through the point $(x_1, y_1, z_1)$ & parallel to the vector $(ai+bj+ck)$ is given as $$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$$
Hence, the equation of the line passing through the point $(1, 0, -3)$ & parallel to the vector $(2i-4j+5k)$ is given as $$\frac{x-1}{2}=\frac{y-0}{-4}=\frac{z-(-3)}{5}$$
$$\frac{x-1}{2}=\frac{-y}{4}=\frac{z+3}{5}$$ It can also be represented as
$$r(t)=(1, 0, -3)+t(2, -4, 5)$$
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
In this case we can use the projection formula $$\operatorname{proj}_{b}a=\frac {a\cdot b}{\|b\|^2}b.$$
Let $L_A=\{(-1,0,1)+t\,(1,1,1)\,|\,t\in \mathbb{R}\}$ be the line passing through $A$ with direction vector $(1,1,1)$
Also, $\vec{AV}=(0,3,2)-(-1,0,1)=(1,3,1)$.
So, $$\vec{AP}=\operatorname{proj}_{(1,1,1)}\vec{AV}=\frac {\vec{AV}\cdot (1,1,1)}{\|(1,1,1)\|^2}(1,1,1)$$ $$=\frac{(1,3,1)\cdot (1,1,1)}{3}(1,1,1)$$ $$\Rightarrow \vec{AP}=\frac{5}{3}(1,1,1).$$