A particle is moving along a straight line parallel to $x$-axis with constant velocity. Find angular momentum about the origin in vector form

$A$ paritcle falls freely near the surface of the earth. Consider $a$ fixed point $O$ (not vertically below the particle) on the ground.

A thin rod of mass $M$ and length $a$ is free to rotate in horizontal plane about a fixed vertical axis passing through point $O$. A thin circular disc of mass $M$ and of radius $a / 4$ is pivoted on this rod with its center at a distance $a / 4$ from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4 \Omega$. The total angular momentum of the system about the point $O$ is $\left(\frac{ M a^2 \Omega}{48}\right) n$. The value of $n$ is. . . . .

$A$ particle of mass $m$ is rotating in a plane is $a$ circular path of radius $r$, its angular momentum is $L$. The centripital force acting on the particle is :

A particle of mass $m$ moves along line $PC$ with velocity $v$ as shown.  What is the angular momentum of the particle about $O$ ​