Solving equation

Let $D$ be a point inside a triangle $\triangle {ABC}$ such that

$|\overline {DA}|=a, |\overline {DB}|=\dfrac{b}{\sqrt 3}, |\overline {DC}|=c, |\overline {AB}|=5, |\overline {BC}|=3, |\overline {CA}|=4, \angle {ADB}=150\degree, \angle {BDC}=90\degree, \angle {CDA}=120\degree$ .

Then $a, b, c$ satisfy the given equations.

Then the sum of the areas of $\triangle {DAB}, \triangle {DBC}, \triangle {DCA}$ is the area of $\triangle {ABC}$ . That is,

$\dfrac{1}{2}\left (ac\sin 120\degree+\frac{ab}{\sqrt 3}\sin 150\degree+\frac{bc}{\sqrt 3}\sin 90\degree\right )=\dfrac{1}{2}\times 3\times 4\implies \dfrac{ab+2bc+3ca}{\sqrt 3}=\boxed {24}$ .