Let $$a, b, c \in R$$ be such that $$a^2 + b^2 + c^2 = 1$$. If $$a\cos\theta = b\cos\left(\theta + \frac{2\pi}{3}\right) = c\cos\left(\theta + \frac{4\pi}{3}\right)$$, where $$\theta = \frac{\pi}{9}$$, then the angle between the vectors $$a\hat{i} + b\hat{j} + c\hat{k}$$ and $$b\hat{i} + c\hat{j} + a\hat{k}$$ is:

We begin by denoting the two vectors

$$$\vec{u}=a\hat i+b\hat j+c\hat k,\qquad \vec{v}=b\hat i+c\hat j+a\hat k.$$$

The angle between two vectors is obtained from the well-known formula

$$\cos\phi=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}|\,|\vec{v}|}.$$

Hence our first task is to evaluate the dot product $$\vec{u}\cdot\vec{v}$$.

We are told that

$$$a\cos\theta=b\cos\!\left(\theta+\frac{2\pi}{3}\right)= c\cos\!\left(\theta+\frac{4\pi}{3}\right),\qquad\theta=\frac{\pi}{9}.$$$ For convenience set

$$k=a\cos\theta=b\cos\!\left(\theta+\frac{2\pi}{3}\right)= c\cos\!\left(\theta+\frac{4\pi}{3}\right).$$

From this single constant $$k$$ we can express each coefficient:

$$$a=\frac{k}{\cos\theta},\qquad b=\frac{k}{\cos\!\left(\theta+\frac{2\pi}{3}\right)},\qquad c=\frac{k}{\cos\!\left(\theta+\frac{4\pi}{3}\right)}.$$$

Now we write the dot product explicitly:

$$\vec{u}\cdot\vec{v}=ab+bc+ca.$$ Substituting the expressions found above gives

$$$\vec{u}\cdot\vec{v}=k^{2}\!\left[ \frac{1}{\cos\theta\,\cos\!\left(\theta+\frac{2\pi}{3}\right)}+ \frac{1}{\cos\!\left(\theta+\frac{2\pi}{3}\right)\cos\!\left(\theta+\frac{4\pi}{3}\right)}+ \frac{1}{\cos\!\left(\theta+\frac{4\pi}{3}\right)\cos\theta}\right].$$$

For brevity let us introduce the three cosine values

$$$p=\cos\theta,\qquad q=\cos\!\left(\theta+\frac{2\pi}{3}\right),\qquad r=\cos\!\left(\theta+\frac{4\pi}{3}\right).$$$

In this notation we have

$$\vec{u}\cdot\vec{v}=k^{2}\!\left(\frac1{pq}+\frac1{qr}+\frac1{rp}\right).$$

The bracketed sum can be rewritten by bringing the terms to a common denominator:

$$\frac1{pq}+\frac1{qr}+\frac1{rp}=\frac{r+p+q}{pqr}.$$

Thus the dot product reduces to

$$\vec{u}\cdot\vec{v}=k^{2}\frac{p+q+r}{pqr}.$$

To proceed we need the value of the numerator $$p+q+r$$. A trigonometric identity for three angles that differ by $$120^\circ$$ (or $$\frac{2\pi}{3}$$) is

$$\cos\theta+\cos\!\left(\theta+\frac{2\pi}{3}\right)+ \cos\!\left(\theta+\frac{4\pi}{3}\right)=0.$$ This result can be checked either by the sum-to-product formulas or more elegantly by writing each cosine as the real part of an exponential and using the fact that $$1+e^{i\frac{2\pi}{3}}+e^{i\frac{4\pi}{3}}=0$$. Therefore

$$p+q+r=0.$$

Substituting this zero into the expression for the dot product gives

$$\vec{u}\cdot\vec{v}=k^{2}\frac{0}{pqr}=0.$$

Because $$a^{2}+b^{2}+c^{2}=1$$, the vector $$\vec{u}$$ is non-zero; cyclic symmetry immediately implies $$\vec{v}$$ is also non-zero. Hence $$|\vec{u}|\,|\vec{v}| \neq 0$$, and the cosine of the required angle is

$$$\cos\phi=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}|\,|\vec{v}|} =\frac{0}{|\vec{u}|\,|\vec{v}|}=0.$$$

An angle whose cosine is $$0$$ is $$\displaystyle\frac{\pi}{2}$$ radians (or $$90^\circ$$).

Hence, the correct answer is Option C.