All the real values of $p, q$ so that the system of equations

$$ 2 x+p y+6 z=8, x+2 y+q z=5 $$

and $\quad x+y+3 z=4$

may have no solution are

If $p$ and $q$ are two distinct real values of $\lambda$ for which the system of equations

$$ \begin{array}{r} (\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0 \\ (\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0 \\ 2 x+(3 \lambda+1) y+3(\lambda-1) z=0 \end{array} $$

has non-zero solution, then $p^2+q^2-p q=$