Given the curve:
\[
y = ax^2 + bx + c
\]
1) Since the tangent at \( (1,1) \) is parallel to the x-axis, the slope at \( x=1 \) must be zero. The slope is given by the derivative:
\[
\frac{dy}{dx} = 2ax + b
\]
At \( x=1 \):
\[
0 = 2a(1) + b = 2a + b \implies b = -2a
\]
2) The point \( (1,1) \) lies on the curve:
\[
1 = a(1)^2 + b(1) + c = a + b + c
\]
3) The point \( (2,3) \) lies on the curve:
\[
3 = a(2)^2 + b(2) + c = 4a + 2b + c
\]
Substitute \( b = -2a \) into the above two equations:
From 2):
\[
1 = a + (-2a) + c = -a + c \implies c = 1 + a
\]
From 3):
\[
3 = 4a + 2(-2a) + c = 4a - 4a + c = c
\]
Therefore,
\[
c = 3
\]
From the earlier expression for \( c \):
\[
c = 1 + a = 3 \implies a = 2
\]
Then,
\[
b = -2a = -4
\]
Finally, calculate \( a + b + c \):
\[
a + b + c = 2 + (-4) + 3 = 1
\]