Revival Of This Problem. Part 2

In triangle PQR as seen in the image, there is a arbitrary point S inside it such that, $\angle PSQ = \angle PSR = \angle QSR = 120^\circ$ .

Now, applying law of cosines, we have:

$a^2 + b^2 - 2ab(\cos 120^\circ) = (\overline{PQ})^2 = (\sqrt{2})^2 \\ b^2 + c^2 - 2bc(\cos 120^\circ) = (\overline{QR})^2 = 1^2 \\ c^2 + a^2 - 2ac(\cos 120^\circ) = (\overline{PR})^2 = (\sqrt{3})^2$

Hence, $\overline{PQ} = \sqrt{2}, \ \overline{QR} = 1, \ \overline{PR} = \sqrt{3}$ and by pythagorean theorem, we can confirm that triangle PQR is a right triangle and hence, Area of Triangle PQR = $\cfrac{\sqrt{2}}{2}$ .

Now, applying the area of a triangle using sines,

$\rightarrow \left( \cfrac{1}{2}\times ab\times (\sin 120^\circ)\right ) + \left( \cfrac{1}{2}\times bc\times (\sin 120^\circ)\right ) + \left( \cfrac{1}{2}\times ca\times (\sin 120^\circ)\right ) = \cfrac{\sqrt{2}}{2} \\ ab + bc + ca = \cfrac{ \sqrt{2} } { \frac{ \sqrt{3}}{2} } = \cfrac{ 2 \sqrt{6} } { 3} = \dfrac{x \sqrt{y}}{z} \\ \implies x + y + z = 11.$