14.7. (PFTB, 68). Find the ratios between the sides of a triangle satisfying the equality

$$
\frac{a \cos \alpha+b \cos \beta+c \cos \gamma}{a \sin \beta+b \sin \gamma+c \sin \alpha}=\frac{p}{9 R}
$$

where $a, b, c$ are the sides of the triangle, $\alpha, \beta, \gamma$ are the measures of the opposite angles, $P$ is the perimeter, and $R$ is the radius of the circumscribed circle.
14.7. Let $h_{a}, h_{b}, h_{c}$ be the altitudes of the triangle satisfying the given condition, and $S$ be its area, then (Fig. 96)

$$
a \sin \beta=h_{c}, \quad b \sin \gamma=h_{a}, \quad c \sin \alpha=h_{b}
$$

and the original equality is equivalent to the condition

$$
P\left(h_{a}+h_{b}+h_{c}\right)=9 R(a \cos \alpha+b \cos \beta+c \cos \gamma)
$$

On one hand, according to Theorem 75, we have $9 R(a \cos \alpha+b \cos \beta+c \cos \gamma)=$

$=9 R(2 R \sin \alpha \cos \alpha+2 R \sin \beta \cos \beta+2 R \sin \gamma \cos \gamma)=$

$$
=9 R^{2}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma)=18 S
$$

On the other hand, using the theorem of means, we get $P\left(h_{a}+h_{b}+h_{c}\right)=(a+b+c)\left(h_{a}+h_{b}+h_{c}\right) \geqslant$

$$
\geqslant 9 \sqrt[3]{a b c} \sqrt[3]{h_{a} h_{b} h_{c}}=9 \sqrt[3]{a h_{a} b h_{b} c h_{c}}=9 \sqrt[3]{(2 S)^{3}}=18 S
$$

Equality is achieved if and only if

$$
a=b=c, \quad h_{a}=h_{b}=h_{c}
$$

i.e., when the triangle is equilateral.