Why is angular momentum not a scalar quantity?

$\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v}$

= $\left(3\widehat{i}-\widehat{j}\right)\times \left(3\widehat{j}+\widehat{k}\right)$

$=9\widehat{k}+3\left(-\widehat{j}\right)-\widehat{i}$

$=-\widehat{i}-3\widehat{j}+9\widehat{k}$

$\left|\overrightarrow{L}\right|=\sqrt[]{1+9+81}=\sqrt[]{91}$

L = Iω
I = (mL²/3) + (mL²/3) + [m (√2L)²/12] + m (L/√2)² = mL² [2/3 + 1/6 + 1/2] = (4mL²/3); L = (4/3)mL²ω

Yes, an object moving in a straight path can also possess angular momentum with respect to a referenc epoint which doesn't lie on it's path. In such cases. the referenc epoint may lie perpendicular distance from the axis. If an object is moving in a straight line (not through the reference point) the perpendicular distance $r_\perp$ (l = rp) will be non-zero.

Yes, how much time these planets take in completing an orbit around the sun hugely depends on their distance from the sun as well as their mass. Since these planets move in a circular orbit, the term angular momentum is used in such cases for computing relevant information.

Using conservation of Angular momentum along axis of rotation, we can write
Mr²ω = (Mr² + 2mr²)ω? ⇒ ω? = Mω / (M + 2m)