If the value of the determinant | array lll a & 1 & 1 1 & ~b & 1 1...

$$\begin{aligned} & \left|\begin{array}{lll}\mathrm{a} & 1 & 1 \\ 1 & \mathrm{ b} & 1 \\ 1 & 1 & \mathrm{c}\end{array}\right| > 0 \\ & \Rightarrow \mathrm{a}(\mathrm{bc}-1)-1(\mathrm{c}-1)+1(1-\mathrm{b}) > 0 \\ & \Rightarrow \mathrm{abc}-\mathrm{a}-\mathrm{c}+1+1-\mathrm{b} > 0 \\ & \Rightarrow \mathrm{abc}+2-(\mathrm{a}+\mathrm{b}+\mathrm{c}) > 0\end{aligned}$$\Rightarrow \mathrm{abc} > (\mathrm{a}+\mathrm{b}+\mathrm{c})-2$Let;$\mathrm{a}=-1 ; \mathrm{b}=0 \ \text{\text{ and } } \mathrm{c}=1$Then;$0 > -2$[which is correct] Hence,$a b c=0\therefore$ After considering all the option; (2) is correct option.