Step-by-step Solution:

1. Find the values of a and b

We are given the matrix M and its adjugate, adj M. $M = \left[ {\begin{matrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \\ \end{matrix} } \right]$ and $adj M = \left[ {\begin{matrix} { - 1} & 1 & { - 1} \\ 8 & { - 6} & 2 \\ { - 5} & 3 & { - 1} \\ \end{matrix} } \right]$

The adjugate of a matrix is the transpose of its cofactor matrix. Let $C_ij$ be the cofactor of the element $m_ij$ in matrix M. Then $(adj M)_ij = C_ji$.

• The element at position (1,1) of adj M is $(adj M)_11 = C_11$. $C_11 = (2)(1) - (3)(b) = 2 - 3b$. From the given adj M, $(adj M)_11 = -1$. So, $2 - 3b = -1 \implies 3b = 3 \implies b = 1$.

• The element at position (2,3) of adj M is $(adj M)_23 = C_32$. $C_32 = -[(0)(3) - (a)(1)] = a$. From the given adj M, $(adj M)_23 = 2$. So, a = 2.

Let's verify these values with another element.

• The element at position (1,2) of adj M is $(adj M)_12 = C_21$. $C_21 = -[(1)(1) - (a)(b)] = ab - 1$. From the given adj M, $(adj M)_12 = 1$. So, $ab - 1 = 1 \implies ab = 2$. Substituting a = 2 and b = 1, we get (2)(1) = 2, which is consistent.

Thus, we have found a = 2 and b = 1.

2. Evaluate Option D: a + b = 3

Using the values we found: a + b = 2 + 1 = 3. Therefore, Option D is correct.

3. Calculate the determinant of M

Now that we have a and b, the matrix M is: $M = \left[ {\begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \\ \end{matrix} } \right]$ We can calculate its determinant: det(M) = 0(2*1 - 3*1) - 1(1*1 - 3*3) + 2(1*1 - 2*3) det(M) = 0 - 1(1 - 9) + 2(1 - 6) det(M) = -(-8) + 2(-5) = 8 - 10 = -2.

Alternatively, we can use the property M(adj M) = det(M)I. The diagonal elements of this product will be det(M). Let's compute the (1,1) element: $(M(adj M))_11 = (0)(-1) + (1)(8) + (a)(-5) = 8 - 5a$. Since a=2, $(M(adj M))_11 = 8 - 5(2) = 8 - 10 = -2$. So, det(M) = -2.

4. Evaluate Option A: $det(adj M^2) = 81$

We use the property $det(adj A) = (det A)^(n-1)$ for an n x n matrix A. Here, $A = M^2$ and n = 3. $det(adj M^2) = (det(M^2))^(3-1) = (det(M^2))^2$. Also, $det(M^2) = (det M)^2$. So, $det(adj M^2) = ((det M)^2)^2 = (det M)^4$. Substituting det M = -2: $det(adj M^2) = (-2)^4 = 16$. Since $16 \neq 81$, Option A is incorrect.

5. Evaluate Option B: If $M[α β γ]^T = [1 2 3]^T$, then α - β + γ = 3

We have the system MX = B, where $X = [α β γ]^T$ and $B = [1 2 3]^T$. The solution is $X = M^-1 B$. We know that $M^-1 = (1/det M) * adj M$. $X = \frac{1}{-2} \left[ {\begin{matrix} { - 1} & 1 & { - 1} \\ 8 & { - 6} & 2 \\ { - 5} & 3 & { - 1} \\ \end{matrix} } \right] \left[ {\begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} } \right]$ $X = \left[ {\begin{matrix} α \\ β \\ γ \\ \end{matrix} } \right] = -\frac{1}{2} \left[ {\begin{matrix} (-1)(1) + (1)(2) + (-1)(3) \\ (8)(1) + (-6)(2) + (2)(3) \\ (-5)(1) + (3)(2) + (-1)(3) \\ \end{matrix} } \right]$ $X = -\frac{1}{2} \left[ {\begin{matrix} -1+2-3 \\ 8-12+6 \\ -5+6-3 \\ \end{matrix} } \right] = -\frac{1}{2} \left[ {\begin{matrix} -2 \\ 2 \\ -2 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 1 \\ -1 \\ 1 \\ \end{matrix} } \right]$ So, α = 1, β = -1, γ = 1. Now, we check the condition: α - β + γ = 1 - (-1) + 1 = 1 + 1 + 1 = 3. This matches the given condition. Therefore, Option B is correct.

6. Evaluate Option C: $(adj M)^-1 + adj(M^-1) = -M$

We use two key properties of adjugate and inverse matrices:

1. $adj(A^-1) = (adj A)^-1$
2. $(adj A)^-1 = A / det(A)$

Using property 1, the left-hand side (LHS) of the equation becomes: $LHS = (adj M)^-1 + (adj M)^-1 = 2(adj M)^-1$.

Now, using property 2: LHS = 2 * (M / det(M)).

We calculated det(M) = -2. Substituting this value: LHS = 2 * (M / -2) = -M.

The given equation is LHS = -M. Since we have shown that LHS = -M, the equation is true. Therefore, Option C is correct.

Conclusion: The correct options are B, C, and D.