⚡ AZUCATION SHORTCUT:
Observe the term $(b - c)$ repeating in both equations.
Let $(b - c) = k$. Equations become $a - 6k = 4$ and $6a + 3k = 50$.
Solving these gives $a = 8$ and $k = 2/3$.
The required value is $2a + 3k = 2(8) + 3(2/3) = 16 + 2 = \mathbf{18}$.

Step 1: Simplify using substitution
Notice that $b$ and $c$ always appear as a pair with opposite signs. Let's group them:
$a - 6(b - c) = 4 \quad \dots(1)$
$6a + 3(b - c) = 50 \quad \dots(2)$
Let $k = b - c$.

Step 2: Solve the Linear System
The equations are now:
(1) $a - 6k = 4$
(2) $6a + 3k = 50$
Multiply equation (2) by $2$ to align the $k$ terms:
$12a + 6k = 100 \quad \dots(3)$
Add equation (1) and equation (3):
$(a - 6k) + (12a + 6k) = 4 + 100$
$13a = 104 \implies \mathbf{a = 8}$.

Step 3: Find the value of k
Substitute $a = 8$ into equation (1):
$8 - 6k = 4 \implies 6k = 4 \implies \mathbf{k = \frac{4}{6} = \frac{2}{3}}$.

Step 4: Calculate the final expression
We need to find the value of $2a + 3b - 3c$, which is $2a + 3(b - c)$ or $2a + 3k$:
Value $= 2(8) + 3\left(\frac{2}{3}\right)$
Value $= 16 + 2 = \mathbf{18}$.

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