So, you have three points,
$$\begin{cases}
A = ( x_A ,\, y_A ,\, z_A ) \\
B = ( x_B ,\, y_B ,\, z_B ) \\
C = ( x_C ,\, y_C ,\, z_C ) \end{cases}$$
The angle $\theta = \angle ABC$ fulfills
$$\cos(\theta) = \frac{ \overline{BA} \cdot \overline{BC} }{\left\lVert \overline{BA} \right\rVert \left\lVert \overline{BC} \right\rVert }$$
and
$$\sin(\theta) = \frac{ \left\lVert \overline{BA} \times \overline{BC} \right\rVert}{\left\lVert \overline{BA} \right\rVert \left\lVert \overline{BC} \right\rVert}$$
where $\cdot$ denotes vector dot product, and $\times$ denotes vector cross product.
If we want to rotate $C$ around $B$ so that the angle $\angle ABC$ becomes $\theta + \varphi$, we first find the normal of the plane the three points form:
$$N = ( x_N ,\, y_N ,\, z_N ) = \frac{\overline{BA} \times \overline{BC}}{\left\lVert\overline{BA} \times \overline{BC}\right\rVert}$$
If we first calculate the cross product components,
$$\begin{cases}
x = ( y_A - y_B ) ( z_C - z_B ) - ( z_A - z_B ) ( y_C - y_B ) \\
y = ( z_A - z_B ) ( x_C - x_B ) - ( x_A - x_B ) ( z_C - z_B ) \\
z = ( x_A - x_B ) ( y_C - y_B ) - ( y_A - y_B ) ( x_C - x_B )
\end{cases}$$
we get the components of the normal vector $N$ by "normalizing" the length of the resulting vector to unit length, or 1:
$$\begin{cases}
x_N = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \\
y_N = \frac{y}{\sqrt{x^2 + y^2 + z^2}} \\
z_N = \frac{z}{\sqrt{x^2 + y^2 + z^2}} \end{cases}$$
We now simply need to rotate $\overline{BC}$ around the vector $N$ by angle $\varphi$.
We can do that by applying Rodrigues' rotation formula to $\overline{BC} = ( x_C - x_B , y_C - y_B , z_C - z_B )$, to get the new position $C'$ relative to $B$. Essentially,
$$C' = B + \overline{BC} \cos(\varphi) + \left( N \times \overline{BC} \right) \sin(\varphi) + N \left( N \cdot \overline{BC} \right) \left(1 - \cos(\varphi) \right)$$
Note that the definition of $\theta$ (and the rotation angle $\varphi$) above define it as positive counterclockwise. If $\angle ABC$ is clockwise — that is, $A$ is to the right of $C$, when viewing from $B$ —, then $\theta$ is negative, and a positive $\varphi$ will then decrease the angle.
If you want to rotate both $A$ and $C$ by the same amount, so that the angle $\angle ABC$ changes from $\theta$ to $\theta + \varphi$, you rotate $A$ by $-\varphi/2$ and $C$ by $\varphi/2$:
$$A' = B + \overline{BA} \cos\left(\frac{\varphi}{2}\right) - \left( N \times \overline{BA} \right) \sin\left(\frac{\varphi}{2}\right) + N \left( N \cdot \overline{BA} \right) \left(1 - \cos\left(\frac{\varphi}{2}\right) \right)$$
because $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$; and
$$C' = B + \overline{BC} \cos\left(\frac{\varphi}{2}\right) + \left( N \times \overline{BC} \right) \sin\left(\frac{\varphi}{2}\right) + N \left( N \cdot \overline{BC} \right) \left(1 - \cos\left(\frac{\varphi}{2}\right) \right)$$
