⚡ AZUCATION SHORTCUT:
Notice $(b-c)$ appears together. Let $(b-c) = x$.
(1) $a - 6x = 4$
(2) $6a + 3x = 50 \implies 12a + 6x = 100$.
Add the two: $13a = 104 \implies a = 8$.
Substitute $a$: $8 - 6x = 4 \implies 6x = 4 \implies 3x = 2$.
Find $2a + 3x$: $2(8) + 2 = \mathbf{18}$. (Correction: Re-evaluating Eq 2).

Step 1: Simplify Variables
Observe that in all expressions, $b$ and $c$ appear as $(b - c)$ or $(3b - 3c)$. Let $(b - c) = k$.
Equation 1: $a - 6k = 4$
Equation 2: $6a + 3k = 50$

Step 2: Solve the Linear System
Multiply Eq 2 by 2: $12a + 6k = 100$.
Add this to Eq 1:
$(a - 6k) + (12a + 6k) = 4 + 100$
$13a = 104 \implies \mathbf{a = 8}$.

Step 3: Find the value of $3k$
Substitute $a = 8$ into Eq 2:
$6(8) + 3k = 50 \implies 48 + 3k = 50 \implies \mathbf{3k = 2}$.

Step 4: Final Calculation
We need to find $2a + 3b - 3c$, which is $2a + 3(b - c) = 2a + 3k$.
Value $= 2(8) + 2 = 16 + 2 = \mathbf{18}$.
(Note: Checking the arithmetic, Option A is 18).

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