When a particle is moving in a circular path around a fixed point in a plane,the direction of its angular momentum is ........

$A$ mass $m$ is moving with a constant velocity $v$ along a line parallel to the $X$-axis. Its angular momentum with respect to the origin or the $Z$-axis is:

$A$ particle of mass $m$ moves along a straight line $AB$ in the $XY$ plane. Let $L_A$ and $L_B$ be the angular momentum of the particle about the origin $O$ when it is at points $A$ and $B$ respectively. Then:

Three equal masses $m$ are kept at vertices $(A, B, C)$ of an equilateral triangle of side $a$ in free space. At $t = 0$,they are given an initial velocity $\vec{V}_A = V_0 \hat{u}_{AC}, \vec{V}_B = V_0 \hat{u}_{BA}$ and $\vec{V}_C = V_0 \hat{u}_{CB}$. Here,$\hat{u}_{AC}, \hat{u}_{CB}$ and $\hat{u}_{BA}$ are unit vectors along the edges of the triangle. If the three masses interact gravitationally,then the magnitude of the net angular momentum of the system about the centroid of the triangle is:

$A$ particle of mass $m$ moves with a constant velocity $\vec{v}$ in the $X-Y$ plane. Its angular momentum with respect to the origin is:

$A$ ring of mass $10 \ kg$ and diameter $0.4 \ m$ is rotated about its axis. If it rotates at $2100 \ rpm$,its angular momentum will be ........... $kg \cdot m^2/s$.