<s>[INST] In a triangle ABC, $\begin {vmatrix} a&b&c \\b&c&a \\c&a&b \end {vmatrix}=0$, then find $\sin A\sin B+\sin B\sin C+\sin C\sin A$ From the given determinant,
$$a^3+b^3+c^3-3abc=0$$
Which implies $a+b+c=0$ or $a^2+b^2+c^2 -ab-bc-ac=0$
Since the former isn’t possible,
$$a^2+b^2+c^2=ab+bc+ac$$
$$\sin A\sin B+\sin B\sin C +\sin C\sin A=\sin^2A +\sin^2B+\sin ^2C$$
The answer for this is either $\cos^2A+\cos^2 B+\cos^2C$ or one of numbers $0, 1, \frac 94$
How do I proceed from here?
 [/INST] The sides $a,b,c$ of the triangle satisfy
$$ a^2+b^2+c^2=ab+bc+ac \\
\begin{matrix} \implies \dfrac{1}{2}(2a^2+2b^2+2c^2-2ab-2bc-2ac) & =0 \\ 
(a-b)^2+(b-c)^2+(c-a)^2 &=0
\end{matrix}$$
This straightaway means that $a=b=c$ (i.e. and equilateral triangle) otherwise the inequality cannot hold.
Therefore, $\sin A\sin B+\sin B\sin C+\sin C\sin A=\dfrac{3}{4}*3=\dfrac{9}{4}$
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