Using only a simple calculator, find the values of $k$ such that $\operatorname{det}(M)=0$, where $M=\left[\begin{array}{ccc}-6 & k & 0 \\ 1 & 1 & k \\ 1 & 1 & -2\end{array}\right]$. As your answer, enter the SUM of the value(s) of $k$ that satisfy this condition.

Step 1 ：We are given a 3x3 matrix M = \[\begin{bmatrix} -6 & k & 0 \\ 1 & 1 & k \\ 1 & 1 & -2 \end{bmatrix}\]

Step 2 ：We need to find the values of k such that the determinant of M is zero. The determinant of a 3x3 matrix is given by the formula: det(M) = a(ei − fh) − b(di − fg) + c(dh − eg), where the matrix M is given by: M = \[\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\]

Step 3 ：Substituting the values from our given matrix into the determinant formula, we get: det(M) = -6(-2 - k) - k(1 - k) + 0

Step 4 ：Solving this equation for k, we find that the values of k that satisfy the condition det(M) = 0 are k = -6 and k = -2

Step 5 ：The sum of these values is -6 + -2 = -8

Step 6 ：Final Answer: \(\boxed{-8}\)