Solving 2-degree equations in 3 variables. We are given 3 equations:
$x^2+\sqrt3 xy + y^2 = 25$
$y^2 + z^2 = 9$
$x^2 +xz+ z^2 = 16$.
$x,y,z$ are positive real numbers.
Then we have to find value of $xy + 2yz + \sqrt3 xz$.
 A: Based on system of equations being sides of a right angled triangle and a point P
inside the triangle such that -
$\angle BPC = 90^0, \angle APC = 150^0, \angle APB = 120^0 $
and, $AP = x, CP = y, BP = z$
We know area of a triangle is $\frac{1}{2} \times$ length of side 1 $\times$ length of side 2 $\times \sin \theta$
where $\theta$ is the angle between side 1 and side 2.
Now sum of area, $\triangle APC + \triangle BPC + \triangle APB = \triangle XYZ$
$\frac{1}{2}(xy\sin150^0 + yz\sin90^0 + xz\sin 120^0) = \frac{1}{2} \times 3 \times 4$
$xy \times \frac{1}{2} + yz + xz \times \frac{\sqrt3}{2} = 12$
$xy + 2yz + \sqrt3 xz = 24$
A: Note that you have $$\forall x,y,z \in \mathbb{R}^{+}: \left\{\begin{aligned} x^{2}+\sqrt{3} xy + y^2 &=& 25\\
y^{2} + z^{2} &=& 9\\
x^{2} +xz+ z^{2} &=& 16  \end{aligned} \right.$$ if, and only if, $$ \forall x,y,z \in \mathbb{R}^{+}: \left\{\begin{aligned} x^{2}+\sqrt{3} xy + y^2 &=& \color{blue}{5}^{2}\\
y^{2} + z^{2} &=& \color{blue}{3}^{2}\\
x^{2} +xz+ z^{2} &=& \color{blue}{4}^{2}  \end{aligned} \right. $$
Now, we can approach this problem as an algebraic geometry problem. Indeed, consider a triangle $\bigtriangleup XYZ$ with side lengths $3,4,5$ and draw a point $P$ inside the triangle such that $XP=x$, $YP=y$, and $ZP=z$. Now, you can considerer he equations in the context of the law of cosines.
Can you continue from here?
A: Here is an accurate picture. Draw in some extra lines......

A: Given a triangle with sides $l_1,l_2,l_3$ we have
$$
\cases{
l_1^2=l_2^2+l_3^2-2l_2l_3\cos\theta_1\\
l_2^2=l_1^2+l_3^2-2l_1l_3\cos\theta_2\\
l_2^2=l_1^2+l_2^2-2l_1 l_2\cos\theta_3
}
$$
then making $l_1=x,l_2=y,l_3=z$
$$
\cases{
2\cos\theta_1=0\\
2\cos\theta_2=-1\\
2\cos\theta_3=-\sqrt{3}
}
$$
It is a rectangle. Etc.
