Answer
The angular momentum about the point is $~~(720~kg~m^2/s)~\hat{k}$
Work Step by Step
We can write the general form of the cross-product:
$a\times b = (a_xb_y-b_xa_y)~\hat{k}$
We can express the particle's position relative to the point $(-2.0~m,-2.0~m)$ in unit-vector notation:
$r = (5.0~m)~\hat{i}+ (-2.0~m)~\hat{j}$
We can express the velocity in unit-vector notation:
$v = (30.0~m/s)~\hat{i}+(60.0~m/s)~\hat{j}$
We can find the angular momentum about the point:
$L = m~(r \times v)$
$L = (2.0~kg)~[(5.0~m)(60.0~m/s)-(30.0~m/s)(-2.0~m)]~\hat{k}$
$L = (2.0~kg)~(360~m^2/s)~\hat{k}$
$L = (720~kg~m^2/s)~\hat{k}$
The angular momentum about the point is $~~(720~kg~m^2/s)~\hat{k}$
