Answer
(a) See explanations.
(b)$-46$
(c) No.
(d) Yes.
Work Step by Step
(a) To test if $(-1,0,1)$ is a solution, plug-in the numbers into each equation:
$-1+2(0)+6(1)=5$,
$-3(-1)-6(0)+5(1)=8$, and
$2(-1)+6(0)+9(1)=7$
Thus, $(-1,0,1)$ is a solution to the equation.
(b) Set up the coefficient determinant and do the operations $R_3+3R_1\to R_3$ and $R_3-2R_1\to R_3$
$\begin{array}( \\|A|= \\ \\ \end{array}
\begin{vmatrix} 1&2&6\\-3&-6&5\\2&6&9 \end{vmatrix} \begin{array}( \\= \\ \\ \end{array}
\begin{vmatrix} 1&2&6\\0&0&23\\0&2&-3 \end{vmatrix}$
Expand the determinant using the first column:
$|A|=(1)(0-23\times2)=-46$
(c) As $|A|\ne0$, the solution above should be unique and there are no other solutions.
(d) Yes, the Cramer's Rule can be used to solve the system as $|A|\ne0$ and the solution is unique.
