Geometric System

Let us draw a triangle $\triangle {ABC}$ right angled at $A$ with $|\overline {AB}|=3, |\overline {AC}|=4, |\overline {BC}|=5$ . Let us take a point $D$ within the triangle such that $\angle {ADB}=90\degree, \angle {ADC}=120\degree, \angle {BDC}=150\degree$ . Let $|\overline {BD}|=x, |\overline {AD}|=y, |\overline {CD}|=z$ . Then $x, y, z$ satisfy the given equations. Now, area of $\triangle {ADB}=\dfrac{1}{2}xy\sin 90\degree=\dfrac{1}{2}xy$ , of $\triangle {ADC}=\dfrac{1}{2}yz\sin 120\degree=\dfrac{\sqrt 3}{4}yz$ , and of $\triangle {BDC}=\dfrac{1}{2}zx\sin 150\degree=\dfrac{1}{4}zx$ . Sum of the areas of these three triangles is the area of $\triangle {ABC}=\dfrac{1}{2}\times 3\times 4=6$ , that is, $\dfrac{xy}{2}+\dfrac{\sqrt 3yz}{4}+\dfrac{zx}{4}=6\implies 2xy+zx+\sqrt 3yz=6\times 4=\boxed {24}$ .