If the system of linear equations : $x+y+2z=6$,$2x+3y+az=a+1$,$-x-3y+bz=2b$ where $a, b \in R$,has infinitely many solutions,then $7a+3b$ is equal to :

The system of equations $4x + y + 2z = 5$,$x - 5y + 3z = 10$,and $9x - 3y + 7z = 20$ has

The sum of three numbers is $6$. Thrice the third number when added to the first number gives $7$. On adding three times the first number to the sum of the second and third numbers,we get $12$. The product of these numbers is:

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).

Solve the system of the following equations: $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$,$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$,and $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$.

If $A = \begin{bmatrix} -1 & 2 \\ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$,$AX = B$,then $X = $