Answer
The angular momentum about the origin is:
$(42.0~kg~m^2/s)~\hat{i}+(24.0~kg~m/s^2)~\hat{j}+(60.0~kg~m^2/s)~\hat{k}$
Work Step by Step
We can write the general form of a cross-product:
$a\times b = (a_yb_z-b_ya_z)~\hat{i}+(a_zb_x-b_za_x)~\hat{j}+(a_xb_y-b_xa_y)~\hat{k}$
We can express the particle's position in unit-vector notation:
$d = (2.00~m)~\hat{i}+ (4.00~m)~\hat{j}- (3.00~m)~\hat{k}$
We can express the velocity in unit-vector notation:
$v = (-6.00~m/s)~\hat{i}+(3.00~m/s)~\hat{j}+(3.00~m/s)~\hat{k}$
We can find the angular momentum about the origin:
$L = m~(d \times v)$
$L = (2.00~kg)~[(4.00~m)(3.00~m/s)-(3.00~m/s)(-3.00~m)]~\hat{i}+[(-3.00~m)(-6.00~m/s)-(3.00~m/s)(2.00~m)]~\hat{j}+[(2.00~m)(3.00~m/s)-(-6.00~m/s)(4.00~m)]~\hat{k}$
$L = (2.00~kg)~[(21.0~m^2/s)~\hat{i}+(12.0~m/s^2)~\hat{j}+(30.0~m^2/s)~\hat{k}]$
$L = (42.0~kg~m^2/s)~\hat{i}+(24.0~kg~m/s^2)~\hat{j}+(60.0~kg~m^2/s)~\hat{k}$
The angular momentum about the origin is:
$(42.0~kg~m^2/s)~\hat{i}+(24.0~kg~m/s^2)~\hat{j}+(60.0~kg~m^2/s)~\hat{k}$
