We have,
$\begin{array}{r}
2 x+p y+6 z=8 \\
x+2 y+q z=5 \\
x+y+3 z=4
\end{array}$
Now, if we express above equation in matrix form, We have,
$A X=B$
Where, $A=\left[\begin{array}{lll}2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3\end{array}\right], B=\left[\begin{array}{l}8 \\ 5 \\ 4\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$
For no solution $|A|=0$
$\begin{aligned}
& \Rightarrow & \left|\begin{array}{lll}
2 & p & 6 \\
1 & 2 & q \\
1 & 1 & 3
\end{array}\right|=0 \\
\Rightarrow & 2(6-q)-p(3-q)+6(1-2) =0 \\
\Rightarrow & 12-2 q-3 p+p q-6 =0 \\
\Rightarrow & 3 p+2 q-p q-6 =0 \\
\Rightarrow & 3 p-6+2 q-p q =0 \\
\Rightarrow & 3(p-2)-q(p-2) =0 \\
\Rightarrow & (p-2)(3-q) =0 \Rightarrow p=2,3
\end{aligned}$
But for $p=2$ equations $2 x+p y+6 z=8$ and $x+y+3 z=4$ are same.
So, $p \neq 2, q=3$