Given system of three equations
\(\begin{aligned}
2 x+3|y|+5[z] & =0 \\
x+|y|-2[z] & =4 \\
x+|y|+[z] & =1
\end{aligned}\)
and
According to Cramer's rule,
\(x=\frac{\Delta_1}{\Delta},|y|=\frac{\Delta_2}{\Delta} \text { and }[z]=\frac{\Delta_3}{\Delta}\)
where,
\(\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
2 & 3 & 5 \\
1 & 1 & -2 \\
1 & 1 & 1
\end{array}\right| \\
& =2(1+2)-3(1+2)+5(1-1)=-3 \\
\Delta_1 & =\left|\begin{array}{ccc}
0 & 3 & 5 \\
4 & 1 & -2 \\
1 & 1 & 1
\end{array}\right| \\
& =0(1+2)-3(4+2)+5(4-1) \\
& =-18+15=-3 \\
\Delta_2 & =\left|\begin{array}{ccc}
2 & 0 & 5 \\
1 & 4 & -2 \\
1 & 1 & 1
\end{array}\right| \\
& =2(4+2)-0(1+2)+5(1-4)=-3
\end{aligned}\)
and \(\Delta_3=\left|\begin{array}{lll}2 & 3 & 0 \\ 1 & 1 & 4 \\ 1 & 1 & 1\end{array}\right|=2(l-4)-3(l-4)+0(l-1)=3\)
Now, \(x=\frac{-3}{-3}=1,|y|=\frac{-3}{-3}=1\) and \([z]=\frac{-3}{3}=-1\)
\(\therefore x=1,|y|=1 \Rightarrow y= \pm 1\) and \([z]=-1\)
\(\Rightarrow z \in[-1,0)\)
So, the given system of three equations has infinitely many solution. Hence, option (3) is correct.