If the system of equations $x+2y-3z=2$,$2x+\lambda y+5z=5$,$14x+3y+\mu z=33$ has infinitely many solutions,then $\lambda+\mu$ is equal to:

Let $\alpha$ be a solution of $x^2+x+1=0$,and for some $a$ and $b$ in $\mathbb{R}$,$\begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$. If $\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$,then $m + n$ is equal to

If the following system of linear equations
$2x + y + z = 5$
$x - y + z = 3$
$x + y + az = b$
has no solution,then :

Let the system of linear equations
$-x+2y-9z=7$
$-x+3y-7z=9$
$-2x+y+5z=8$
$-3x+y+13z=\lambda$
has a unique solution $x=\alpha, y=\beta, z=\gamma$. Then the distance of the point $(\alpha, \beta, \gamma)$ from the plane $2x-2y+z=\lambda$ is

Let $A$ and $B$ be two $3 \times 3$ non-zero real matrices such that $AB$ is a zero matrix. Then:

All the real values of $p, q$ so that the system of equations $\begin{cases} 2x + py + 6z = 8 \\ x + 2y + qz = 5 \\ x + y + 3z = 4 \end{cases}$ may have no solution are