For two $$3 \times 3$$ matrices $$A$$ and $$B$$, let $$A + B = 2B'$$ and $$3A + 2B = I_3$$, where $$B'$$ is the transpose of $$B$$ and $$I_3$$ is $$3 \times 3$$ identity matrix. Then:

We have two relations between the unknown square matrices $$A$$ and $$B$$ of order $$3$$:

$$A + B = 2B' \qquad\qquad (1)$$

$$3A + 2B = I_3 \qquad\qquad (2)$$

The symbol $$B'$$ denotes the transpose of $$B$$ and $$I_3$$ is the identity matrix of order $$3$$.

From relation (1) isolate $$A$$ in terms of $$B$$ and its transpose:

$$A = 2B' - B \qquad\qquad (3)$$

Substitute the expression (3) for $$A$$ in relation (2):

$$3(2B' - B) + 2B \;=\; I_3$$

$$6B' - 3B + 2B \;=\; I_3$$

$$6B' - B \;=\; I_3 \qquad\qquad (4)$$

Because the transpose of the identity matrix is itself, we may take the transpose of both sides of (4) without changing the right-hand side:

$$(6B' - B)' \;=\; I_3'$$

$$6B - B' \;=\; I_3 \qquad\qquad (5)$$

Now equations (4) and (5) give a pair of linear matrix equations in the two unknowns $$B$$ and $$B'$$. Solve them simultaneously. From (4) express $$B$$:

$$B = 6B' - I_3 \qquad\qquad (6)$$

Insert (6) into (5):

$$6(6B' - I_3) - B' \;=\; I_3$$

$$36B' - 6I_3 - B' \;=\; I_3$$

$$35B' \;=\; 7I_3$$

$$B' \;=\; \dfrac{7}{35}I_3 \;=\; \dfrac{1}{5}I_3 \qquad\qquad (7)$$

Transpose (7) to obtain $$B$$ itself (since the transpose of $$I_3$$ is $$I_3$$):

$$B \;=\; \left(B'\right)' \;=\; \dfrac{1}{5}I_3 \qquad\qquad (8)$$

Return to (3) to determine $$A$$:

$$A = 2B' - B$$

Using (7) and (8):

$$A = 2\!\left(\dfrac{1}{5}I_3\right) - \dfrac{1}{5}I_3 = \dfrac{2}{5}I_3 - \dfrac{1}{5}I_3 = \dfrac{1}{5}I_3 \qquad\qquad (9)$$

Thus both matrices are fixed uniquely:

$$A = \dfrac{1}{5}I_3, \qquad B = \dfrac{1}{5}I_3$$

Now evaluate each given option.

Option A:

$$10A + 5B = 10\!\left(\dfrac{1}{5}I_3\right) + 5\!\left(\dfrac{1}{5}I_3\right) = 2I_3 + I_3 = 3I_3$$

Option B:

$$3A + 6B = 3\!\left(\dfrac{1}{5}I_3\right) + 6\!\left(\dfrac{1}{5}I_3\right) = \dfrac{3}{5}I_3 + \dfrac{6}{5}I_3 = \dfrac{9}{5}I_3 \neq 2I_3$$

Option C:

$$5A + 10B = 5\!\left(\dfrac{1}{5}I_3\right) + 10\!\left(\dfrac{1}{5}I_3\right) = I_3 + 2I_3 = 3I_3 \neq 2I_3$$

Option D:

$$B + 2A = \dfrac{1}{5}I_3 + 2\!\left(\dfrac{1}{5}I_3\right) = \dfrac{1}{5}I_3 + \dfrac{2}{5}I_3 = \dfrac{3}{5}I_3 \neq I_3$$

Only Option A produces the identity relation demanded by the option statement.

Hence, the correct answer is Option A.