Consider the following system of equations: $x+2y-3z=a$,$2x+6y-11z=b$,and $x-2y+7z=c$,where $a, b$,and $c$ are real constants. Then the system of equations:

$A$ trust fund has Rs. $30,000$ that must be invested in two different types of bonds. The first bond pays $5 \%$ interest per year,and the second bond pays $7 \%$ interest per year. Using matrix multiplication,determine how to divide Rs. $30,000$ among the two types of bonds if the trust fund must obtain an annual total interest of Rs. $1800$.

Let $\alpha, \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations:
$x+2y+z=7$
$x+\alpha z=11$
$2x-3y+\beta z=\gamma$
Match each entry in List-$I$ to the correct entries in List-$II$:

List-$I$	List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has	$(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has	$(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has	$(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has	$(4)$ $x=11, y=-2$ and $z=0$ as a solution
	$(5)$ $x=-15, y=4$ and $z=0$ as a solution

If $\begin{bmatrix} 2 & 1 & 1 \\ 0 & 3 & -1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$, then $\begin{bmatrix} x \\ y \\ z \end{bmatrix} =$

If the system of simultaneous linear equations $x+y+z=\lambda$,$5x-y+\mu z=10$,and $2x+3y-z=6$ has a unique solution,then:

If the system of equations $\alpha x + y + z = 5$,$x + 2y + 3z = 4$,and $x + 3y + 5z = \beta$ has infinitely many solutions,then the ordered pair $(\alpha, \beta)$ is equal to: