Step 1: For a system of linear equations to have no solution, the equations must be inconsistent. This generally happens when:
\[
\text{Rank of coefficient matrix} \neq \text{Rank of augmented matrix}
\]
Step 2: Consider the coefficient matrix:
\[
A =
\begin{bmatrix}
2 & p & 6 \\
1 & 2 & q \\
1 & 1 & 3
\end{bmatrix}
\quad \text{and} \quad
B =
\begin{bmatrix}
2 & p & 6 & 8 \\
1 & 2 & q & 5 \\
1 & 1 & 3 & 4
\end{bmatrix}
\]
Step 3: We aim to choose \( p \) and \( q \) such that one of the rows in the augmented matrix leads to an inconsistency after row reduction. Testing multiple values, we find that the system becomes inconsistent (no solution) when:
\[
p \neq 2, \quad \text{and} \quad q = 3
\]
For these values, the last row in the row-reduced form turns into something like:
\[
0x + 0y + 0z = c \quad \text{where} \quad c \neq 0
\]
indicating inconsistency.