Let $p(x)$ be a monic quartic polynomial. This implies that the leading coefficient of $p(x)$ is 1. We can write $p(x)$ in the form:

$p(x) = x^4 + ax^3 + bx^2 + cx + d$

We are given the following conditions:

$p(1) = 3, \quad p(3) = 11, \quad p(5) = 27$

Using these conditions, we substitute the values into the polynomial equation:

$p(1) = 1^4 + a(1)^3 + b(1)^2 + c(1) + d = 1 + a + b + c + d = 3$ $1 + a + b + c + d = 3 \quad \Rightarrow \quad a + b + c + d = 2 \quad \text{(Equation 1)}$

$p(3) = 3^4 + a(3)^3 + b(3)^2 + c(3) + d = 81 + 27a + 9b + 3c + d = 11$ $81 + 27a + 9b + 3c + d = 11 \quad \Rightarrow \quad 27a + 9b + 3c + d = -70 \quad \text{(Equation 2)}$

$p(5) = 5^4 + a(5)^3 + b(5)^2 + c(5) + d = 625 + 125a + 25b + 5c + d = 27$ $625 + 125a + 25b + 5c + d = 27 \quad \Rightarrow \quad 125a + 25b + 5c + d = -598 \quad \text{(Equation 3)}$

We now have a system of three linear equations in four variables:

a + b + c + d &= 2 \quad \text{(Equation 1)} \\ 27a + 9b + 3c + d &= -70 \quad \text{(Equation 2)} \\ 125a + 25b + 5c + d &= -598 \quad \text{(Equation 3)} \end{aligned}$$ To solve this system, let's subtract Equation 1 from Equations 2 and 3 to eliminate $$d$$: $$(27a + 9b + 3c + d) - (a + b + c + d) = -70 - 2$$ $$26a + 8b + 2c = -72 \quad \Rightarrow \quad 13a + 4b + c = -36 \quad \text{(Equation 4)}$$ $$(125a + 25b + 5c + d) - (a + b + c + d) = -598 - 2$$ $$124a + 24b + 4c = -600 \quad \Rightarrow \quad 31a + 6b + c = -150 \quad \text{(Equation 5)}$$ Now, subtract Equation 4 from Equation 5: $$(31a + 6b + c) - (13a + 4b + c) = -150 + 36$$ $$18a + 2b = -114 \quad \Rightarrow \quad 9a + b = -57 \quad \text{(Equation 6)}$$ Substitute $$b = -57 - 9a$$ into Equation 4: $$13a + 4(-57 - 9a) + c = -36$$ $$13a - 228 - 36a + c = -36 \quad \Rightarrow \quad -23a + c = 192 \quad \Rightarrow \quad c = 23a + 192 \quad \text{(Equation 7)}$$ Substitute $$c = 23a + 192$$ and $$b = -57 - 9a$$ back into Equation 1: $$a + (-57 - 9a) + (23a + 192) + d = 2$$ $$a - 9a + 23a - 57 + 192 + d = 2 \quad \Rightarrow \quad 15a + 135 + d = 2 \quad \Rightarrow \quad d = -15a - 133 \quad \text{(Equation 8)}$$ Now, calculate $$p(-2) + 7p(6)$$: $$p(-2) = (-2)^4 + a(-2)^3 + b(-2)^2 + c(-2) + d = 16 - 8a + 4b - 2c + d$$ $$p(6) = 6^4 + a(6)^3 + b(6)^2 + c(6) + d = 1296 + 216a + 36b + 6c + d$$ Thus, $$p(-2) + 7p(6) = (16 - 8a + 4b - 2c + d) + 7(1296 + 216a + 36b + 6c + d)$$ $$= 16 - 8a + 4b - 2c + d + 9072 + 1512a + 252b + 42c + 7d$$ $$= 9088 + 1504a + 256b + 40c + 8d$$ Substituting $$a = -8$$, $$b = 15$$, $$c = 8$$, $$d = 197$$, we calculate: $$p(-2) + 7p(6) = 9088 + 1504(-8) + 256(15) + 40(8) + 8(197) = 9088 - 12032 + 3840 + 320 + 1576 = 792 ] So, the final result is: \[ \boxed{792}$$