Answer
(a) $2700000\sqrt 3$
(b) $4677$ liters
Work Step by Step
First we identify the given vectors as: $\vec w=\langle 300, 0, 0 \rangle$, $\vec u=\langle 0, 120, 0 \rangle$, and $\vec v=\langle 150sin30^{\circ}, 0, 150cos30^{\circ} \rangle=\langle 75, 0, 75\sqrt 3 \rangle$
(a) The cross product is $\vec v\times\vec w=\langle (0\times0- 75\sqrt 3 \times0), ( 75\sqrt 3 \times300-75\times0), (75\times0-0\times300) \rangle=\langle 0, 22500\sqrt 3, 0 \rangle$, and the scalar triple product is
$\vec u\cdot(\vec v\times\vec w)=0\times0+120\times22500\sqrt 3+0\times0=2700000\sqrt 3$
(b) As $1L=1000cm^3$ and the unit of the above number is $cm^3$, the capacity of the tank is the volume of the
parallelepiped calculated above, $V=2700000\sqrt 3/1000\approx4677L$
