The set of equations $x - y + 3z = 2$,$2x - y + z = 4$,and $x - 2y + \alpha z = 3$ has:

The number of real values of $\lambda$ for which the system of linear equations $2x + 4y - \lambda z = 0$,$4x + \lambda y + 2z = 0$,and $\lambda x + 2y + 2z = 0$ has infinitely many solutions is:

Let $A=\begin{bmatrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6 \end{bmatrix}$ and $B=[b_{ij}]_{3 \times 3}$ with $b_{11}=2, b_{13}=-2, b_{12}=0$ such that $AB=\begin{bmatrix} 2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12 \end{bmatrix}$. Then $|B|+\operatorname{trace}(B)=$

If the system of equations $2x + 9y + 5z = 8$,$2x + 3y - z = -4$,$x - 2z = -5$ has an infinite number of solutions $x = -5 + at$,$y = 2 + bt$,$z = ct$,$t \in R$,then $a$,$b$,$c$ respectively are

The equations $x+y+z=3$,$x+2y+2z=6$ and $x+ay+3z=b$ have

The system of linear equations $\lambda x + y + z = 3$,$x - y - 2z = 6$,and $-x + y + z = \mu$ has: