It does not violate angular momentum because sphere A has angular momentum about the origin that equals
$$ \boldsymbol{L}_A = \boldsymbol{r}_A \times \boldsymbol{p}_A $$
where $\boldsymbol{p}_A = m_A \boldsymbol{v}_A$ is (translational) momentum of A and $\boldsymbol{v}_A$ is the velocity vector of A.
After the first collision, if sphere A is motionless, the entirety of this angular momentum is transferred into sphere B
$$ \boldsymbol{L}_A = \boldsymbol{L}_B = \boldsymbol{r}_B \times \boldsymbol{p}_B $$
And later, then two spheres B and C start to interact, via an exchange of impulses along the string, and because these impulses are offset from the coordinate system, they also exchange some of their angular momentum such that
$$ \boldsymbol{L}_B^{\text{before}} = \boldsymbol{L}_B^{\text{after}} + \boldsymbol{L}_C^{\text{after}} $$
where again
$$\begin{aligned} \boldsymbol{L}_B^{\text{after}} = \boldsymbol{r}_B \times \boldsymbol{p}_B^{\text{after}} \\ \boldsymbol{L}_C^{\text{after}} = \boldsymbol{r}_C \times \boldsymbol{p}_C^{\text{after}} \\ \end{aligned}$$
you can confirm this since the momentum of sphere C is purely due to the impulse from the string
$$\begin{aligned} \boldsymbol{p}_B^{\text{after}} = \boldsymbol{p}_B^{\text{before}} - \boldsymbol{J}_{\text{string}} \\ \boldsymbol{p}_C^{\text{after}} = + \boldsymbol{J}_{\text{string}} \\ \end{aligned} $$
If you do an accounting of momentum you get
$$\begin{array}{c|lll|c} \text{Momentum} & \text{Sphere A} & \text{Sphere B} & \text{Sphere C} & \text{Total Momentum}\\ \hline \text{Stage 1} & \boldsymbol{p}_{A}=m_{A}\boldsymbol{v}_{A} & \boldsymbol{p}_{B}=0 & \boldsymbol{p}_{C}=0 & m_{A}\boldsymbol{v}_{A}\\ \text{Stage 2} & \boldsymbol{p}_{A}=m_{A}\boldsymbol{v}_{A}-\boldsymbol{J}_{AB} & \boldsymbol{p}_{B}=\boldsymbol{J}_{AB} & \boldsymbol{p}_{C}=0 & m_{A}\boldsymbol{v}_{A}\\ \text{Stage 3} & \boldsymbol{p}_{A}=m_{A}\boldsymbol{v}_{A}-\boldsymbol{J}_{AB} & \boldsymbol{p}_{B}=\boldsymbol{J}_{AB}-\boldsymbol{J}_{BC} & \boldsymbol{p}_{C}=\boldsymbol{J}_{BC} & m_{A}\boldsymbol{v}_{A} \end{array}$$
Each $\boldsymbol{J}_{AB}$ and $\boldsymbol{J}_{BC}$ represents in impulse vector describing the impact, and the tightening of the string.
And more importantly an accounting of angular momentum
$$\begin{array}{c|lll|c} \text{Angular Mom.} & \text{Sphere A} & \text{Sphere B} & \text{Sphere C} & \text{Total Ang. Mom.}\\ \hline \text{Stage 1} & \boldsymbol{L}_{A}=\boldsymbol{r}_{A}\times\boldsymbol{p}_{A} & \boldsymbol{L}_{B}=0 & \boldsymbol{L}_{C}=0 & \boldsymbol{r}_{A}\times\boldsymbol{p}_{A}\\ \text{Stage 2} & \boldsymbol{L}_{A}=\boldsymbol{r}_{A}\times\left(\boldsymbol{p}_{A}-\boldsymbol{J}_{AB}\right) & \boldsymbol{L}_{B}=\boldsymbol{r}_{B}\times\boldsymbol{J}_{AB} & \boldsymbol{L}_{C}=0 & \boldsymbol{r}_{A}\times\boldsymbol{p}_{A}\\ \text{Stage 3} & \boldsymbol{L}_{A}=\boldsymbol{r}_{A}\times\left(\boldsymbol{p}_{A}-\boldsymbol{J}_{AB}\right) & \boldsymbol{L}_{B}=\boldsymbol{r}_{B}\times\left(\boldsymbol{J}_{AB}-\boldsymbol{J}_{BC}\right) & \boldsymbol{L}_{C}=\boldsymbol{r}_{C}\times\boldsymbol{J}_{BC} & \boldsymbol{r}_{A}\times\boldsymbol{p}_{A} \end{array}$$
Note the condition that $\left(\boldsymbol{r}_{A}-\boldsymbol{r}_{B}\right)\times\boldsymbol{J}_{AB}=0$ which requires the impulse $\boldsymbol{J}_{AB}$ to be inline with the line connecting the centers of sphere A and B. Similarly, $\left(\boldsymbol{r}_{B}-\boldsymbol{r}_{C}\right)\times\boldsymbol{J}_{BC}=0$ for the second impulse $\boldsymbol{J}_{BC}$ due to the string.