1. State the problem. We have the system of equations: $-x-y-z=3$ $2x+y+z=-5$ $x+2y+z=0$ We are asked to calculate the determinant of the coefficient matrix. 2. Write the coefficient matrix. The coefficients of $x,y,z$ form the matrix $$A=\begin{pmatrix}-1&-1&-1\\2&1&1\\1&2&1\end{pmatrix}$$ We need $\det(A)$. 3. Use a determinant formula (row 1 expansion). For $A=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$, $$\det(A)=a\begin{vmatrix}e&f\\h&i\end{vmatrix}-b\begin{vmatrix}d&f\\g&i\end{vmatrix}+c\begin{vmatrix}d&e\\g&h\end{vmatrix}$$ Here $a=-1,\; b=-1,\; c=-1,\; d=2,\; e=1,\; f=1,\; g=1,\; h=2,\; i=1$. 4. Compute the $2\times2$ minors. $\begin{vmatrix}e&f\\h&i\end{vmatrix}=\begin{vmatrix}1&1\\2&1\end{vmatrix}=1\cdot1-1\cdot2=1-2=-1$. $\begin{vmatrix}d&f\\g&i\end{vmatrix}=\begin{vmatrix}2&1\\1&1\end{vmatrix}=2\cdot1-1\cdot1=2-1=1$. $\begin{vmatrix}d&e\\g&h\end{vmatrix}=\begin{vmatrix}2&1\\1&2\end{vmatrix}=2\cdot2-1\cdot1=4-1=3$. 5. Substitute into the expansion and simplify carefully. $$\det(A)=(-1)(-1)-(-1)(1)+(-1)(3)$$ $$\det(A)=1+1-3$$ $$\det(A)=-1$$ 6. Final answer. The determinant of the coefficient matrix is $-1$.