Solve the following system of linear equations: 2a + 3b + 4c = 11, a + 5b + 7c = 15, a + 11b + 13 = 25

Subtract 13 from both sides of Equation (3):

$$a + 11b = 25 - 13$$ $$a + 11b = 12 \quad (\text{Equation 3 simplified})$$

Substitute $a = 12 - 11b$ into Equation (1):

$$2(12 - 11b) + 3b + 4c = 11$$ $$24 - 22b + 3b + 4c = 11$$ $$-19b + 4c = 11 - 24$$ $$-19b + 4c = -13 \quad (\text{Equation 5})$$

Substitute $a = 12 - 11b$ into Equation (2):

$$(12 - 11b) + 5b + 7c = 15$$ $$12 - 6b + 7c = 15$$ $$-6b + 7c = 15 - 12$$ $$-6b + 7c = 3 \quad (\text{Equation 6})$$

We have: 5) $-19b + 4c = -13$ 6) $-6b + 7c = 3$

Multiply Equation (5) by 7 and Equation (6) by 4 to eliminate $c$:

$$7(-19b + 4c) = 7(-13) \implies -133b + 28c = -91 \quad (\text{Equation 7})$$ $$4(-6b + 7c) = 4(3) \implies -24b + 28c = 12 \quad (\text{Equation 8})$$

Subtract Equation (8) from Equation (7):

$$(-133b + 28c) - (-24b + 28c) = -91 - 12$$ $$-133b + 24b = -103$$ $$-109b = -103$$ $$b = \frac{-103}{-109}$$ $$b = \frac{103}{109}$$

$$-6\left(\frac{103}{109}\right) + 7c = 3$$ $$-\frac{618}{109} + 7c = 3$$ $$7c = 3 + \frac{618}{109}$$ $$7c = \frac{3 \times 109}{109} + \frac{618}{109}$$ $$7c = \frac{327 + 618}{109}$$ $$7c = \frac{945}{109}$$ $$c = \frac{945}{109 \times 7}$$ $$c = \frac{135}{109}$$

$$a = 12 - 11b$$ $$a = 12 - 11\left(\frac{103}{109}\right)$$ $$a = 12 - \frac{1133}{109}$$ $$a = \frac{12 \times 109}{109} - \frac{1133}{109}$$ $$a = \frac{1308 - 1133}{109}$$ $$a = \frac{175}{109}$$

The solution to the system of equations is:

$$a = \boxed{\frac{175}{109}}$$ $$b = \boxed{\frac{103}{109}}$$ $$c = \boxed{\frac{135}{109}}$$