The value of $[\vec{a}+2\vec{b}-\vec{c},\vec{a}-\vec{b},\vec{a}-\vec{b}-\vec{c}]$ is:

The scalar triple product of vectors $\vec{u}=x_1\vec{a}+y_1\vec{b}+z_1\vec{c}$, $\vec{v}=x_2\vec{a}+y_2\vec{b}+z_2\vec{c}$, and $\vec{w}=x_3\vec{a}+y_3\vec{b}+z_3\vec{c}$ is given by:
$$[\vec{u},\vec{v},\vec{w}]=\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|[\vec{a},\vec{b},\vec{c}]$$
Given:$\vec{u}=1\vec{a}+2\vec{b}-1\vec{c}$$\vec{v}=1\vec{a}-1\vec{b}+0\vec{c}$$\vec{w}=1\vec{a}-1\vec{b}-1\vec{c}$ The determinant is:
$$\Delta =\left|\begin{array}{ccc}1& 2& -1\\ 1& -1& 0\\ 1& -1& -1\end{array}\right|$$
Expand along the first row:
$$\Delta =1(1-0)-2(-1-0)-1(-1+1)$$
$$\Delta =1+2+0=3$$
Thus, the value is $3[\vec{a},\vec{b},\vec{c}]$.