The value of $$\mid M \mid$$ is ..........

The three equations can be written in matrix form as$$\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}= \begin{bmatrix}\alpha\\\beta\\\gamma-1\end{bmatrix}$$

Denote the coefficient matrix by $$A$$ and the right-hand column by $$\mathbf{b}=(\alpha,\beta,\gamma-1)^{T}.$$
Since $$\det A=0,$$ the rank of $$A$$ is $$2.$$ For the system to be consistent, $$\mathbf{b}$$ must satisfy the same linear relation that makes the rows of $$A$$ dependent.

Find that dependence: let $$c_1\,(1,2,3)+c_2\,(4,5,6)+c_3\,(7,8,9)=\mathbf 0.$$ Solving, we obtain $$c_1=1,\;c_2=-2,\;c_3=1,$$ i.e. $$\text{Row}_1-2\,\text{Row}_2+\text{Row}_3=\mathbf 0.$$

Applying this to the right-hand sides gives$$\alpha-2\beta+(\gamma-1)=0\;\;\Longrightarrow\;\;\alpha-2\beta+\gamma=1.$$Hence the set of all consistent triples $$(\alpha,\beta,\gamma)$$ forms the plane

$$P:\;\alpha-2\beta+\gamma-1=0.\qquad-(1)$$

Now evaluate the determinant of the matrix$$M=\begin{bmatrix}\alpha&2&\gamma\\\beta&1&0\\-1&0&1\end{bmatrix}.$$

Expanding along the first row:

$$\begin{aligned}\lvert M\rvert &=\alpha\begin{vmatrix}1&0\\0&1\end{vmatrix}-2\begin{vmatrix}\beta&0\\-1&1\end{vmatrix}+\gamma\begin{vmatrix}\beta&1\\-1&0\end{vmatrix}\\[4pt] &=\alpha(1)-2(\beta)+\gamma(1)\\ &=\alpha-2\beta+\gamma.\qquad-(2)\end{aligned}$$

Using the plane condition $$(1),$$ equation $$(2)$$ gives$$\lvert M\rvert=1.$$

Therefore, the required value is 1, which lies in the interval 0.95 - 1.05 provided.