Let $f\left(x\right)=x^2-14x+52$ and $g\left(x\right)=ax+b$, where $a$ and $b$ are positive. Find $a$, given that $f\left(g\left(-5\right)\right)=3$ and $f\left(g\left(0\right)\right)=103$.

$\text{(A) }2\qquad\text{(B) }5\qquad\text{(C) }7\qquad\text{(D) }10\qquad\text{(E) }17$
1. Given the functions \( f(x) = x^2 - 14x + 52 \) and \( g(x) = ax + b \), we need to find the values of \( a \) and \( b \) such that \( f(g(-5)) = 3 \) and \( f(g(0)) = 103 \).

2. First, we substitute \( g(x) \) into \( f(x) \):
   \[
   f(g(x)) = f(ax + b) = (ax + b)^2 - 14(ax + b) + 52
   \]

3. Given \( f(g(-5)) = 3 \), we substitute \( x = -5 \) into \( g(x) \):
   \[
   g(-5) = a(-5) + b = -5a + b
   \]
   Then, substitute \( g(-5) \) into \( f(x) \):
   \[
   f(-5a + b) = (-5a + b)^2 - 14(-5a + b) + 52 = 3
   \]
   Expanding and simplifying:
   \[
   (-5a + b)^2 - 14(-5a + b) + 52 = 3
   \]
   \[
   25a^2 - 10ab + b^2 + 70a - 14b + 52 = 3
   \]
   \[
   25a^2 - 10ab + b^2 + 70a - 14b + 49 = 0 \quad \text{(Equation 1)}
   \]

4. Given \( f(g(0)) = 103 \), we substitute \( x = 0 \) into \( g(x) \):
   \[
   g(0) = a(0) + b = b
   \]
   Then, substitute \( g(0) \) into \( f(x) \):
   \[
   f(b) = b^2 - 14b + 52 = 103
   \]
   Simplifying:
   \[
   b^2 - 14b + 52 = 103
   \]
   \[
   b^2 - 14b - 51 = 0 \quad \text{(Equation 2)}
   \]

5. Solve Equation 2 for \( b \):
   \[
   b = \frac{14 \pm \sqrt{14^2 - 4 \cdot 1 \cdot (-51)}}{2 \cdot 1}
   \]
   \[
   b = \frac{14 \pm \sqrt{196 + 204}}{2}
   \]
   \[
   b = \frac{14 \pm \sqrt{400}}{2}
   \]
   \[
   b = \frac{14 \pm 20}{2}
   \]
   \[
   b = 17 \quad \text{(since \( b \) is positive)}
   \]

6. Substitute \( b = 17 \) into Equation 1:
   \[
   25a^2 - 10a(17) + 17^2 + 70a - 14(17) + 49 = 0
   \]
   \[
   25a^2 - 170a + 289 + 70a - 238 + 49 = 0
   \]
   \[
   25a^2 - 100a + 100 = 0
   \]
   \[
   25(a^2 - 4a + 4) = 0
   \]
   \[
   a^2 - 4a + 4 = 0
   \]
   \[
   (a - 2)^2 = 0
   \]
   \[
   a = 2
   \]

The final answer is \( \boxed{2} \).