<s>[INST] Find the value of $ab+ 2cb+\sqrt3 ac$? Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$,  $b^2+c^2=9$  and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$?
Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find $a,b,c$        
 [/INST] Here is a geometrical solution:

by rewriting the equations as
$$\left\{ 
\begin{aligned}
5^2&=a^2+b^2-2ab\cos\frac{5\pi}{6}\\
3^2&=b^2+c^2-2bc\cos\frac{\pi}{2}\\
4^2&=c^2+a^2-2ca\cos\frac{2\pi}{3}
\end{aligned}
\right.
$$
and evaluating
$$4\left(\frac{1}{2}ab\sin\frac{5\pi}{6}+\frac{1}{2}bc\sin\frac{\pi}{2}+\frac{1}{2}ca\sin\frac{2\pi}{3}\right)=4S_{\Delta ABC}=24.$$

Sketch of a existence proof:
Since $3^2+4^2=5^2$, $\Delta ABC$ exists.  Consider $\overset{\mmlToken{mo}{⏜}}{HC}$, i.e., the segment that lies between $AB$ and $CA$ of a circle, for which $BC$ is a diameter.  For any $P$ on the arc, $\angle BPC=\pi/2$ while $\angle CPA$ increases continuously from $\pi/2$ to $\pi$ (from $H$ to $C$), so there exists $P$ so that $\angle CPA=2\pi/3$.
In general, if such configuration does exist, the point where three segments meet is the intersection of three arcs.  Each arc results the given opening angle with respect to each side of the triangle and they sum up to $2\pi$.
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