If $a+b+c=3, a^{2}+b^{2}+c^{2}=6$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=$ 1 where $a, b$ and $c$ are non-zero numbers, then $a b c=?$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

$\Rightarrow \frac{b c+c a+a b}{a b c}=1$

$\Rightarrow b c+a c+a b=a b c \ldots$ (i)

Again, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+a c)$

$\Rightarrow(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b c$

$\Rightarrow 3^{2}=6+2 a b c$

$\Rightarrow 9=6+2 a b c$

$\Rightarrow 3=2 a b c \Rightarrow \mathrm{abc}=\frac{3}{2}$