The position vectors of three points A , B and C are a , b and c...

Given $\vec{c} = (\cos^2\theta)\vec{a} + (\sin^2\theta)\vec{b}$

Substituting the value of $\vec{c}$ in the second term of the expression $(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})$:

$\vec{b} \times \vec{c} = \vec{b} \times ((\cos^2\theta)\vec{a} + (\sin^2\theta)\vec{b})$

$\vec{b} \times \vec{c} = (\cos^2\theta)(\vec{b} \times \vec{a}) + (\sin^2\theta)(\vec{b} \times \vec{b})$

Since $\vec{b} \times \vec{b} = \vec{0}$ and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$:

$\vec{b} \times \vec{c} = -(\cos^2\theta)(\vec{a} \times \vec{b})$

Substituting the value of $\vec{c}$ in the third term:

$\vec{c} \times \vec{a} = ((\cos^2\theta)\vec{a} + (\sin^2\theta)\vec{b}) \times \vec{a}$

$\vec{c} \times \vec{a} = (\cos^2\theta)(\vec{a} \times \vec{a}) + (\sin^2\theta)(\vec{b} \times \vec{a})$

Since $\vec{a} \times \vec{a} = \vec{0}$:

$\vec{c} \times \vec{a} = -(\sin^2\theta)(\vec{a} \times \vec{b})$

Adding the three terms together:

$(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) = (\vec{a} \times \vec{b}) - (\cos^2\theta)(\vec{a} \times \vec{b}) - (\sin^2\theta)(\vec{a} \times \vec{b})$

$= (\vec{a} \times \vec{b}) [1 - (\cos^2\theta + \sin^2\theta)]$

Using the trigonometric identity $\cos^2\theta + \sin^2\theta = 1$:

$= (\vec{a} \times \vec{b}) [1 - 1]$

$= \vec{0}$

Answer: $\vec{0}$