3. The algebraic expression
$$
\sqrt{x^{2}+6 x+13}+\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}-4 y+5}
$$

reaches its minimum value when $x$ and $y$ are $\qquad$
3. $x=-1, y=1$.

Original algebraic expression
$$
\begin{array}{c}
=\sqrt{(x+3)^{2}+(0+2)^{2}}+ \\
\sqrt{(x-0)^{2}+(0-y)^{2}}+ \\
\sqrt{(0-1)^{2}+(y-2)^{2}} .
\end{array}
$$

As shown in Figure 9, construct the figure with $A(-3,-2) 、 B(x, 0)$ 、 $C(0, y) 、 D(1,2)$.
Then the original algebraic expression $=A B+B C+C D \geqslant A D$.
Equality holds if and only if points $B 、 C$ are on line segment $A D$.

Therefore, the algebraic expression achieves its minimum value if and only if points $B$ 、 $C$ are on line segment $A D$.

Hence, $B 、 C$ are the intersection points of line $A D$ with the
$x$-axis and $y$-axis.
Let the equation of $A D$ be $y=k x+b$. Then
$$
\left\{\begin{array} { l } 
{ - 3 k + b = - 2 , } \\
{ k + b = 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
k=1, \\
b=1 .
\end{array}\right.\right.
$$

Therefore, the equation of $A D$ is $y=x+1$.
When $y=0$, $x=-1$; when $x=0$, $y=1$.
Thus, $B(-1,0) 、 C(0,1)$.
Hence, $x=-1, y=1$.