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 $p^2+q^2-p q=$

Let $z=x+i y$ be a complex number, $A=\{z /|z| \leq 2\}$ and $B=\{z /(1-i) z+(1+i) \bar{z} \geq 4\}$ Then which one of the following options belongs to $A \cap B$ ?

The solutions of the equation $z^2\left(1-z^2\right)=16, z \in \mathbf{C}$, lie on the curve