Let $A = \begin{bmatrix} 1 & 1 & 2 \\ -2 & 0 & 1 \\ 1 & 3 & 5 \end{bmatrix}$. Then the sum of all elements of the matrix $\text{adj}(\text{adj}(2(\text{adj}A)^{-1}))$ is equal to:

The determinant of matrix $A$ is given by:

$|A| = 1(0 - 3) - 1(-10 - 1) + 2(-6 - 0)$

$|A| = -3 + 11 - 12 = -4$

We know the property $A(\text{adj}A) = |A|I$, which gives $\text{adj}A = |A|A^{-1}$.

Taking the inverse on both sides:

$(\text{adj}A)^{-1} = \dfrac{A}{|A|} = \dfrac{A}{-4}$

Let $B = 2(\text{adj}A)^{-1}$. Substituting the above expression, we get:

$B = 2\left(\dfrac{A}{-4}\right) = -\dfrac{1}{2}A$

We need to find the matrix $\text{adj}(\text{adj}B)$. For a square matrix of order $n$, we have $\text{adj}(\text{adj}B) = |B|^{n-2}B$. Since the order is $n = 3$:

$\text{adj}(\text{adj}B) = |B|^{3-2}B = |B|B$

Now, we calculate the determinant of $B$:

$|B| = \left|-\dfrac{1}{2}A\right| = \left(-\dfrac{1}{2}\right)^3|A| = -\dfrac{1}{8}(-4) = \dfrac{1}{2}$

Substituting $|B|$ back into the expression for $\text{adj}(\text{adj}B)$:

$\text{adj}(\text{adj}B) = \dfrac{1}{2}B = \dfrac{1}{2}\left(-\dfrac{1}{2}A\right) = -\dfrac{1}{4}A$

The sum of all elements of matrix $A$ is:

$1 + 1 + 2 - 2 + 0 + 1 + 1 + 3 + 5 = 12$

Therefore, the sum of all elements of the matrix $-\dfrac{1}{4}A$ is:

$-\dfrac{1}{4} \times 12 = -3$

Answer: $-3$