The path followed by the motorist is a regular hexagon with side $500\, m$, as shown in the given figure

Let the motorist start from point $P$. The motorist takes the third turn at $S$.

$\therefore$ Magnitude of displacement $= PS = PV + VS =500+500=1000 \,m$

Total path length $= PQ + QR + RS =500+500+500=1500\, m$

The motorist takes the sixth turn at point $P$, which is the starting point.

$\therefore$ Magnitude of displacement $=0$ Total path length $= PQ + QR + RS + ST + TU + UP$

$=500+500+500+500+500+500=3000 \,m$

The motorist takes the eight turn at point $R$

$\therefore$ Magnitude of displacement $= PR$

$=\sqrt{ PQ ^{2}+ QR ^{2}+2( PQ ) \cdot( QR ) \cos 60^{\circ}}$

$=\sqrt{500^{2}+500^{2}+\left(2 \times 500 \times 500 \times \cos 60^{\circ}\right)}$

$=\sqrt{250000+250000+\left(500000 \times \frac{1}{2}\right)}$

$=866.03\, m$

$\beta=\tan ^{-1}\left(\frac{500 \sin 60^{\circ}}{500+500 \cos 60^{\circ}}\right)=30^{\circ}$

Therefore, the magnitude of displacement is $866.03\, m$ at an angle of $30^{\circ}$ with $PR$. Total path length $=$ Circumference of the hexagon $+ PQ + QR$ $=6 \times 500+500+500=4000\, m$

The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table

Turn	Magnitude of displacement	Total path length
Third	$1000 $	$1500 $
Sixth	$0 $	$3000 $
Eighth	$866.03 ; 30^{\circ}$	$4000$