Given : $r = 7$ cm, angle of minor sector $= 120^\circ$, so angle of major sector $\theta = 360^\circ - 120^\circ = 240^\circ$.

(i) Perimeter of major sector $OACB$
Step 1: Length of major arc $ACB$ :
$$l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{240}{360} \times 2 \times \frac{22}{7} \times 7$$
Step 2: $$l = \frac{2}{3} \times 44 = \frac{88}{3}\ \text{cm}$$
Step 3: Perimeter $=$ major arc $+$ two radii
$$= \frac{88}{3} + 7 + 7 = \frac{88 + 42}{3} = \frac{130}{3} \approx 43{\cdot}33\ \text{cm}$$

(ii) Area of the shaded (minor) segment
Step 4: Area of minor sector $OAB$ (angle $120^\circ$) :
$$= \frac{120}{360} \times \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times 49 = \frac{154}{3} \approx 51{\cdot}33\ \text{cm}^2$$
Step 5: Area of segment $=$ Area of sector $-$ Area of $\triangle OAB$
$$= 51{\cdot}33 - 21{\cdot}2 = 30{\cdot}13\ \text{cm}^2$$
Hence perimeter of major sector $= \dfrac{130}{3}$ cm and area of shaded segment $\approx 30{\cdot}13\ \text{cm}^2$.