The number of $3 \times 3$ matrices $A$ whose entries are either $0$ or $1$ and for which the system $A\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ has exactly two distinct solutions,is

Consider the system of linear equations $a_1x + b_1y + c_1z + d_1 = 0$,$a_2x + b_2y + c_2z + d_2 = 0$ and $a_3x + b_3y + c_3z + d_3 = 0$. Let us denote by $\Delta (a,b,c)$ the determinant $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$. If $\Delta (a,b,c) \neq 0$,then the value of $x$ in the unique solution of the above equations is:

The system of equations $3x + 2y + z = 6$, $3x + 4y + 3z = 14$ and $6x + 10y + 8z = a$ has an infinite number of solutions if $a$ is equal to

If $x^a y^b=e^m, x^c y^d=e^n, \Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|, \Delta_2=\left|\begin{array}{ll}a & m \\ c & n\end{array}\right|, \Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$,then the values of $x$ and $y$ are respectively ($e$ is the base of natural logarithm).

If $p, q, r$ are $3$ real numbers satisfying the matrix equation $[p, q, r] \begin{bmatrix} 3 & 4 & 1 \\ 3 & 2 & 3 \\ 2 & 0 & 2 \end{bmatrix} = [3, 0, 1]$,then $2p + q - r$ equals

Solve the following system of equations by matrix method: $3x - 2y + 3z = 8$,$2x + y - z = 1$,$4x - 3y + 2z = 4$.