In a ΔABC, if $\begin{vmatrix}1&a&b\\1&c&a\\1&b&c\end{vmatrix}=0$, then $\sin^2 A + \sin^2 B+ \sin^2 C$ is

$\frac{9}{4}$

We have,

$\begin{vmatrix}1&a&b\\1&c&a\\1&b&c\end{vmatrix}=0$

$⇒\begin{vmatrix}1&a&b\\0&c-a&a-b\\0&b-a&c-b\end{vmatrix}=0$  [Applying $R_2 → R_2-R_1, R_3 → R_3-R_1$]

$⇒(c-a) (c-b) + (a - b)^2 = 0$

$⇒a^2 + b^2 + c^2 -ab-bc - ca = 0$

$⇒2a^2+2b^2 +2c^2-2ab - 2bc - 2ca = 0$

$⇒(a - b)^2+(b-c)^2 + (c-a)^2 = 0$

$⇒a=b=c$

$⇒ ΔABC$ is equilateral

$⇒A +B+ C =\frac{π}{3}$

$∴ \sin^2 A+ \sin^2 B+ \sin^2 C = 3 \sin^2\frac{π}{3}=\frac{9}{4}$