In a$ ΔABC,$ if $\begin{vmatrix}1 & a & b\\1 & c & a\\1 & b & c\end{vmatrix}=0,$ then

$sin^2A +sin^2B+sin^2C$ is

$\frac{9}{4}$

The correct answer is option (2) : $\frac{9}{4}$

We have,

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

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

$⇒(c-a)(c-b)+(a-b)^2=0$

$⇒a^2+b^2 +c^2-ab-bc-ca=0$

$⇒2a^2 +2b^2 +2c^2 -2ab -2bc -2ca=0$

$⇒(a-b)^2 +(b-c)^2+(c-a)^2=0$

$⇒a=b=c$

$⇒Δ ABC$ is equilateral

$⇒ A=B = C=\frac{\pi }{3}$

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