Solution: The matrix $M$ satisfies the equation \begin{align*} 6M^{-1} = M^2 - 6M + \alpha I & \implies 6I = M^3 - 6M^2 + \alpha M \\ & \implies M^3 - 6M^2 + \alpha M - 6I = 0. \end{align*} Therefore, the characteristic polynomial of the matrix $M$ is \[ x^3 - 6x^2 + \alpha x - 6 = 0. \] Note that the given matrix $M$ is an upper triangular matrix, so the eigenvalues are the diagonal elements, hence the characteristic polynomial of $M$ will be \begin{align*} (x-3)(x-2)(x-1) = 0 & \implies \left( x^2 - 5x + 6 \right) (x - 1) = 0 \\ & \implies x^3 - 6x^2 + 11x - 6 = 0. \end{align*} Therefore, the value of $\alpha$ is equal to $11$.

This can also be solved by the following observation. If $a,b$ and $c$ are the roots of a cubic polynomial then, the polynomial can be written as \[ p(x) = x^3 - \left( a + b + c \right) x^2 + (ab + bc + ac) x - abc. \] Here, the roots are $1,2$ and $3$. So, the characteristic equation will be \[ x^3 - 6x + 11x - 6 = 0, \] and hence $\alpha = 11$.