Q. If for a triangle $A B C\begin{vmatrix}1 & a & b \\ 1 & c & a \\ 1 & b & c\end{vmatrix}=0$ then the value of $\sin ^{2} A +\sin ^{2} B +\sin ^{2} C$ is

KCETKCET 2022 Report Error

$\frac{3 \sqrt{3}}{2}$

$\frac{4}{9}$

$1$

$\frac{9}{4}$

Solution:

$\begin{vmatrix}1 & a & b \\ 1 & c & a \\ 1 & b & c\end{vmatrix}=0$
$1\left(c^{2}-a b\right)-a(c-a)+b(b-c)=0$
$c^{2}-a b-a c+ a^{2}+b^{2}-b c=0 \times b y 2$
$2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a=0$
$a^{2}+b^{2}-2 a b+ b^{2}+c^{2}-2 b c +c^{2}+a^{2}-2 c a=0$
$(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$
$\therefore a-b=0 b-c=0 c-a=0$
$\therefore a=b b=c c-a=0$
$\therefore a=b=c \therefore A=B=C=60^{0}$
$ \sin ^{2} A+\sin ^{2} B+\sin ^{2} C=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}=\frac{9}{4}$

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