GATE Civil Engineering (CE) 2018 Shift 2 Solved Paper

As $\mathit{a},\mathit{b},\mathit{c}$ are non-negative integers and given $\log \mathit{a}+\log \mathit{b}+\log \mathit{c}=0$
$\log \left(\mathit{a}\times \mathit{b}\times \mathit{c}\right)=\log 1$
$\Rightarrow \mathit{a}=\mathit{b}=\mathit{c}=1$
Which can be possible for simple values.
$\mathit{a}=\mathit{b}=\mathit{c}=1$
Hence $\mathit{a}+\mathit{b}+\mathit{c}=1+1+1=3$
Alternate Method:
Given, $\mathit{a}=\mathit{b}=\mathit{c}=1$
$\log \mathit{a}+\log \mathit{b}+\log \mathit{c}=0$
As we know $\log 1=0$, so each one of them can be zero if $\mathit{a}=\mathit{b}=\mathit{c}=1$
$\log 1+\log 1+\log 1=0$
By putting $\mathit{a}=\mathit{b}=1$ equation satisfies
$\mathit{a}+\mathit{b}+\mathit{c}=1+1+1=3$