We have a vector field: $F=x\hat e_x+z\hat e_y+x\hat e_z$
I have to find the line integral:
$\displaystyle\oint_C \boldsymbol F \cdot d \boldsymbol l$
where the contour $C$ consists of: a straight line from the origin to $(0, 0, b)$; a straight line from $(0, 0, b)$ to $(0, b, 0)$; and a straight line from $(0, b, 0)$ to the origin,
There seems to be a few mistakes.
In a vector field (not scalar), the line integral is given by
$I = \displaystyle \int_C \vec{F} \cdot d\vec{r} = \int_C \vec{F}(\vec{r}(t)) \cdot r'(t) \, dt$
a) Straight line from $(0,0,0)$ to $(0,0,b)$ can be parameterized as $\vec{r}(t) = (0,0,b)t \,$ where $0 \leq t \leq 1$.
$x = 0, y = 0, z = bt$
Vector field is $(x,z,x) = (0,bt,0)$
Line integral is $\displaystyle \int_0^1 (0,bt,0) \cdot (0,0,b) \,dt = 0 \,$ as the dot product is zero.
b) Straight line from $(0,0,b)$ to $(0,b,0)$ can be parameterized as $\vec{r}(t) = (0,0,b) + (0,b,-b)t \,$ where $0 \leq t \leq 1$.
$x = 0, y = bt, z = b-bt$
Vector field is $(x,z,x) = (0,b-bt,0)$
Line integral is $\displaystyle \int_0^1 (0,b-bt,0) \cdot (0,b,-b) \,dt = \int_0^1 (b^2-b^2t) \,dt = \frac{b^2}{2}$
Can you do similarly for straight line from $(0,b,0)$ to $(0,0,0)$? (it should be zero)
So your line integral over the given path is $\frac{b^2}{2}$.