Regarding linear momentum operator as vector operator
The vector operator $\hat V$ are defined as the vectors which satisfies the commutator,
$$[\hat L_i,\hat V_j]=i\hbar\epsilon_{ijk}\hat V_k.$$
$\hat L$ is the angular momentum operator.
Thus, if the coordinates are rotated by an angle $\theta$ in anticlockwise direction, then
$\begin{pmatrix}\hat p_x' \ \hat p_y'\ \hat p_z'\end{pmatrix}=\begin{pmatrix} \cos\theta &\sin\theta & 0 \ -\sin\theta & \cos\theta & 0 \ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}\hat p_x \ \hat p_y\ \hat p_z\end{pmatrix}\tag{1}$
But I am not able to prove $(1)$ from the definition of momemtum operator and apply transformation.
After the passive transformation, $\hat p'(x',y',z')=\hat p(x(x',y',z'),y(x',y',z'),z(x',y',z') \ =\hat p_x (x(x',y',z'),y(x',y',z'),z(x',y',z'))\hat x(x',y',z')+ \hat p_y' (x(x',y',z'),y(x',y',z'),z(x',y',z'))\hat y'(x',y',z') + \hat p_z'(x(x',y',z'),y(x',y',z'),z(x',y',z'))\hat z'(x',y',z') \tag{2}$
As $\begin{pmatrix}\hat x' \ \hat y'\ \hat z'\end{pmatrix}=\begin{pmatrix} \cos\theta &\sin\theta & 0 \ -\sin\theta & \cos\theta & 0 \ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}\hat x\ \hat y\ \hat z\end{pmatrix}$
Inverting the above matrix we get $x(x',y',z'),y(x',y',z'),z(x',y',z')$ and so on.
$\begin{pmatrix}\hat x \ \hat y\ \hat z\end{pmatrix}=\begin{pmatrix} \cos\theta &-\sin\theta & 0 \ \sin\theta & \cos\theta & 0 \ 0 & 0 & 1\end{pmatrix} \begin{pmatrix}\hat x'\ \hat y'\ \hat z'\end{pmatrix}$
So, $(2)$ becomes,
$\boxed{\hat p'_x(x',y',z')=\Big(\hat p_x(x(x',y',z'),y(x',y',z'),z(x',y',z'))\cos\theta -\hat p_y(x(x',y',z'),y(x',y',z'),z(x',y',z'))\sin\theta\Big)}$
$\boxed{\hat p'_y(x',y',z')=\Big(\hat p_x(x(x',y',z'),y(x',y',z'),z(x',y',z'))\sin\theta +\hat p_y(x(x',y',z'),y(x',y',z'),z(x',y',z'))\cos\theta\Big)}$
$\boxed{\hat p'_z(x',y',z')= \Big(p_z(x(x',y',z'),y(x',y',z'),z(x',y',z'))\Big)}$
We can see that the form of $(1)$ and the boxed equations differ by the sign of $\sin\theta$ are same. 101