To determine which systems of linear equations have no solution, we examine each provided system:
1. The first system:
[tex]\[
\begin{array}{l}
x + y + z = 1100 \\
x - 2y - z = -500 \\
2x + 3y + 2z = 2600
\end{array}
\][/tex]
2. The second system:
[tex]\[
\begin{array}{l}
x + y + z = 1400 \\
x - 2y - z = -500 \\
2x + 2y + 2z = 2700
\end{array}
\][/tex]
3. The third system:
[tex]\[
\begin{array}{l}
x + y + z = 1900 \\
x - y - 2z = -2000 \\
2x + 2y + z = 1100
\end{array}
\][/tex]
4. The fourth system:
[tex]\[
\begin{array}{l}
x + y + z = 1500 \\
x - y - z = -500 \\
2x + y + z = 2000
\end{array}
\][/tex]
5. The fifth system:
[tex]\[
\begin{array}{l}
x + y + z = 1400 \\
-0.5x - 0.5y - 0.5z = -900 \\
2x + 3y + 2z = 3000
\end{array}
\][/tex]
6. The sixth system:
[tex]\[
\begin{array}{l}
x + y + z = 2400 \\
2x - 2y + 2z = 700 \\
x + 3y + z = 2400
\end{array}
\][/tex]
After solving these systems, we find that only systems 2, 5, and 6 have no solution. Therefore, these are the systems without a solution:
[tex]\[
\begin{array}{l}
x + y + z = 1400 \\
x - 2y - z = -500 \\
2x + 2y + 2z = 2700
\end{array}
\][/tex]
[tex]\[
\begin{array}{l}
x + y + z = 1400 \\
-0.5x - 0.5y - 0.5z = -900 \\
2x + 3y + 2z = 3000
\end{array}
\][/tex]
[tex]\[
\begin{array}{l}
x + y + z = 2400 \\
2x - 2y + 2z = 700 \\
x + 3y + z = 2400
\end{array}
\][/tex]
So, the systems of linear equations that have no solution are the second, fifth, and sixth sets of equations.
