The path followed by the motorist is a regular hexagon with a side 500 m as given in the figure.

Let the motorist start from point P.

The motorist takes the third turn at S.

Therefore,

Magnitude of the displacement = PS = PV + VS

= 500 + 500 = 1000 m

Total path length = PQ + QR + RS

= 500 + 500 + 500 = 1500 m

The motorist takes the 6^th 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.

So, Magnitude of displacement = PR

$=\sqrt{P{Q}^2+Q{R}^2+2PQ\times QR\times \cos {60}^0}$
$=\sqrt{{500}^2+{500}^2+\left(2\times 500\times 500\times \frac{1}{2}\right)}$
$=863.03 m$
$\beta ={\tan }^{-1}\left(\frac{500\sin {60}^0}{500=500\cos {60}^0}\right)={30}^0$
$$Therefore, the magnitude of displacement is 866.03 m at an angle of 30° with PR.

Total path length = Circumference of the hexagon + PQ + QR

= 6 × 500 + 500 + 500 = 4000 m

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