Find the equilateral triangle 🔺

Since $B$ has a smaller $x$ -coordinate, it is near the LEFT .
With some logic reasoning, we get that $B$ is not just on line $l$ , but it is also on line $m$ rotated $60^\circ$ counterclockwise from point $A$ . Think! Why? (Hint: it is on an equilateral triangle, so B rotated 60 degrees clockwise is on m)
This works in the different direction for $C$ , which I will use in my approach.

Now we find that $y=5$ rotated $60$ degrees clockwise from the origin is $-\sqrt3+k$ . Why? (Hint: slope & tangent 60)
Call this line $l’$ . Let’s find $k$ .
We know that the distance between $l’$ and $A$ is $5$ . Construct the line perpendicular to $l’$ at $P$ . So $\overline{AP}=5$ . Solve for $\sqrt{(\frac{1}{\sqrt{3}}n)^2+n^2}=5.$ We get $n=\frac{5\sqrt{3}}{2}$ .

Now we solve for $f(x)=-\sqrt{3}x+k$ with $f(\frac{5\sqrt{3}}{2})=2.5$ .
We find $k=10$ . Now since $C$ is on lines $l’$ & $m$ , we can solve for $C$ : $-\sqrt3x_2+10=3$ $-\sqrt3x_2=-7$ $x_2=\frac{7}{\sqrt{3}}=\frac{7\sqrt{3}}{3}.$ So $C(\frac{7\sqrt{3}}{3},3)$ .
Above pic shows how far we’ve gone now :j (ignore the $B$ )

Now we solve for $B$ by rotating $C$ $60$ degrees counterclockwise from $A$ .
We can use a formula , so $B(x_1,y_1)$ is as follows:
$x_1=\frac{7\sqrt{3}}{3}\cos 60^\circ-3\sin 60^\circ$ ,
$y_1= \frac{7\sqrt{3}}{3}\sin 60^\circ+3\cos 60^\circ$ . So $B=B(-\frac{1}{\sqrt3},5)$ .

So the answer is $\boxed{\frac{32}{\sqrt3}}=\boxed{18.4752...}$
BUT because of my mistake, the answer is $18.289335$ so please subtract the value stated in the question :P