<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 cubic need not be factorised, as side lengths are positive. By AM-GM the triangle is equilateral, leading to @AnindyaPrithvi's calculation of the desired trigonometry.
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