Given equation of the circle is $x^2+y^2+2 x+2 y-3=0$
Which can be written as: $(x+1)^2+(y+1)^2=5$ It is a circle with centre $(-1,-1)$ and radius $\sqrt{5}$ Given line is: $2 x+y+13=0$
To find the required distance, we find the equation of a line perpendicular to the given line, and passing through the centre of the given circle.
$\therefore \quad$ Equation of this line is: $(y+1)=\frac{1}{2}(x+1)$ i.e., $x=2 y+1$
Now, we find the points where line $x=2 y+1$ intersects the circle $x^2+y^2+2 x+2 y-3=0$
$$
\begin{array}{ll}
\therefore & (2 y+1)^2+y^2+2(2 y+1)+2 y-3=0 \\
\therefore & 4 y^2+4 y+1+y^2+4 y+2+2 y-3=0 \\
\therefore & 5 y^2+10 y=0 \\
\therefore & y(y+2)=0 \\
\therefore & y=0 \text { or } y=-2 \\
\therefore & x=1 \text { or } x=-3
\end{array}
$$
$\therefore \quad(1,0)$ and $(-3,-2)$ are the points on the circle, and one of them is at the maximum distance from the given line.
$$
\begin{array}{ll}
\therefore & \mathrm{d}_1=\left|\frac{2(1)+(0)+13}{\sqrt{4+1}}\right| \text { and } \mathrm{d}_2=\left|\frac{2(-3)+(-2)+13}{\sqrt{4+1}}\right| \\
\therefore & \mathrm{d}_1=\frac{15}{\sqrt{5}}=3 \sqrt{5} \quad \text { and } \mathrm{d}_2=\frac{5}{\sqrt{5}}=\sqrt{5} \\
\therefore & \lambda=3 \sqrt{5}
\end{array}
$$