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Well, with my not-so-advanced vector knowledge, I've got a simpler approach in mind.

We name the angle bisector $n$. Let $A(3,0)$ and $P$ be a point in $n$ such that the distance from $P$ to $m$ is $\sqrt{10}$. Drawing the projection of $\vec{AP}$ onto $m$, we get a right triangle and plugging in some trigonometry, we find that $|\vec{AP}|\sin\angle(m,n)=\sqrt{10}$.

How do we calculate the sine? Let's take this problem to the third dimension. Then if we let $\vec b_1$ and $\vec b_2$ be the direction vectors of $m$ and $n$ (respectively), using the cross product we get that $$\sin\angle(m,n)=\sin\angle(\vec b_1,\vec b_2)=\frac{|\vec b_1\times\vec b_2|}{|\vec b_1||\vec b_2|}$$

Note that taking this to the third dimension requires us to rewrite
$$A(3,0,0),\vec b_1=\begin{pmatrix}3\\1\\0\end{pmatrix},\vec b_2=\begin{pmatrix}1\\1\\0\end{pmatrix}$$

Then,
$$\begin{aligned}|\vec{AP}|\frac{|\vec b_1\times\vec b_2|}{|\vec b_1||\vec b_2|}&=\sqrt{10}\\|\vec{AP}|\frac{|2\hat k|}{\sqrt{10}\sqrt2}&=\sqrt{10}\\|\vec{AP}|&=\frac{\sqrt{10}\sqrt2}{2}\sqrt{10}=\boxed{5\sqrt2}\end{aligned}$$

Now, let $k$ be a scalar such that $\vec{AP}=k\vec b_2=\begin{pmatrix}k\\k\\0\end{pmatrix}$. By solving for $|\vec{AP}|=|k\vec b_2|$, we get 2 possible values for $k$ and hence 2 possible coordinates for point $P$. 🙂