I am going to write out a solution here, since some incorrect conclusions were drawn by the OP. The vector field $\overrightarrow{F} = \ $ is in fact conservative, as the curl calculation yields
$\overrightarrow{\nabla} \times \overrightarrow{F} \ = \ < 2xz-2xz\ , \ -2yz - (-2yz) \ , \ z^2 - z^2> \ = \ \overrightarrow{0} . $
So the line integral $\int_C \ (yz^2) dx \ + \ (xz^2) dy \ + \ (2xyz) dz \ $ from $ A(-1,2,-2) $ to $ B(1,5,2) $ is path-independent. $ \\ $
Following the path $C$ specified in the problem statement, we have, for the three segments,
$\int_{C_1} \ (yz^2) dx \ + \ (xz^2) dy \ + \ (2xyz) dz \ \ [dx = 0 \ , \ dy = 0 ]$
$+ \ \int_{C_2} \ (yz^2) dx \ + \ (xz^2) dy \ + \ (2xyz) dz \ \ [dy = 0 \ , \ dz = 0 ]$
$+ \ \int_{C_3} \ (yz^2) dx \ + \ (xz^2) dy \ + \ (2xyz) dz \ \ [dx = 0 \ , \ dz = 0 ]$
$= \ \int_{-2}^2 \ (2xyz) dz \ + \ \int_{-1}^1 \ (yz^2) dx \ + \ \int_{2}^5 \ (xz^2) dy $
$= \ 0^{*} + \ (xyz^2) \ |_{-1, y=2,z=2}^1 \ + \ (xz^2y) \ |_{2, x=1,z=2}^5 $
[with the first integral equal to zero since an odd function of $z$ is being integrated over a path symmetric about the xy-plane]
$= \ [ \ (1 \cdot 2 \cdot 2^2) \ - \ (-1 \cdot 2 \cdot 2^2) \ ] \ + \ [ \ (1 \cdot 2^2 \cdot 5) \ - \ (1 \cdot 2^2 \cdot 2) \ ] $
$= \ \ (8) \ - \ (-8) \ + \ (20) \ - \ (8) \ = \ 28 \ . \\ $
Because the integral is path-independent, we can verify this result by choosing a path $C'$ directly from $A$ to $B$ along the parametrized line segment
$x \ = \ -1 + 2t \ \ [dx = 2 dt] \ , \ y \ = \ 2 + 3t \ \ [dy = 3 dt] \ , \ z \ = \ -2 + 4t \ \ [dz = 4 dt] \ , $
over the range $t: \ 0 \rightarrow 1 \ .$ Our line integral becomes
$\int_0^1 \ (2 + 3t) \cdot (-2 + 4t)^2 \cdot 2 \ dt \ + \ (-1 + 2t) \cdot (-2 + 4t)^2 \cdot 3 \ dt \ $
$+ \ 2 \cdot (-1 + 2t) \cdot (2 + 3t) \cdot (-2 + 4t) \cdot 4 \ dt \ $
[passing over a lot of tedious algebra]
$= \ \int_0^1 \ 36 \ - \ 48t \ - \ 240t^2 \ + \ 384t^3 \ dt \ = \ ( 36t \ - \ 24t^2 \ - \ 80t^3 \ + \ 96t^4 ) \ |_0^1 \ = \ 28 \ . $