Two sequences [a, b, c] and [A, B, C] are related as, \[\left[ {\begin{array}{*{20}{c}} A \\ B \\ C \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{W_3^{ – 1}}&{W_3^{ – 2}} \\ 1&{W_3^{ – 2}}&{W_3^{ – 4}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a \\ b \\ c \end{array}} \right]\] where, $${W_3} = {e^{i\frac{{2\pi }}{3}}}.$$ If another sequence [p, q, r] is derived as, \[\left[ {\begin{array}{*{20}{c}} a \\ b \\ c \end{array}} \right] = \] \[\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{W_3^1}&{W_3^2} \\ 1&{W_3^2}&{W_3^4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{W_3^2}&0 \\ 0&0&{W_3^4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {A/3} \\ {B/3} \\ {C/3} \end{array}} \right]\] then the relationship between the sequences [p, q, r] and [a, b, c] is

= [b, a, c]” option2=”[p, q, r] = [b, c, a]” option3=”[p, q, r] = [c, a, b]” option4=”[p, q, r] = [c, b, a]” correct=”option1″]

The correct answer is $\boxed{\text{(C)}}$.

Let’s start by simplifying the given equation. We can factor out a $1/3$ from the third column of the matrix on the right-hand side, so we get:

$$\left[ {\begin{array}{{20}{c}} a \ b \ c \end{array}} \right] = \frac{1}{3} \left[ {\begin{array}{{20}{c}} 1&1&1 \ 1&{W_3^1}&{W_3^2} \ 1&{W_3^2}&{W_3^4} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 1 \ {W_3^2} \ {W_3^4} \end{array}} \right]\left[ {\begin{array}{{20}{c}} {A} \ {B} \ {C} \end{array}} \right]$$

Now, let’s define a new matrix $M$ as follows:

$$M = \frac{1}{3} \left[ {\begin{array}{{20}{c}} 1&1&1 \ 1&{W_3^1}&{W_3^2} \ 1&{W_3^2}&{W_3^4} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 1 \ {W_3^2} \ {W_3^4} \end{array}} \right]$$

Then, we can rewrite the equation as follows:

$$\left[ {\begin{array}{{20}{c}} a \ b \ c \end{array}} \right] = M \left[ {\begin{array}{{20}{c}} {A} \ {B} \ {C} \end{array}} \right]$$

This means that the sequences [p, q, r] and [a, b, c] are related by the matrix $M$. In other words, if we multiply the sequence [a, b, c] by the matrix $M$, we get the sequence [p, q, r].

Now, let’s look at the options. Option A says that [p, q, r] = [b, a, c]. This is not possible, because the first element of [p, q, r] is $p$, which is equal to $a$. However, the first element of [b, a, c] is $b$. Therefore, option A is not correct.

Option B says that [p, q, r] = [b, c, a]. This is also not possible, because the second element of [p, q, r] is $q$, which is equal to $b$. However, the second element of [b, c, a] is $c$. Therefore, option B is also not correct.

Option C says that [p, q, r] = [c, a, b]. This is possible, because the first element of [p, q, r] is $p$, which is equal to $a$. The second element of [p, q, r] is $q$, which is equal to $b$. And the third element of [p, q, r] is $r$, which is equal to $c$. Therefore, option C is correct.

Option D says that [p, q, r] = [c, b, a]. This is not possible, because the third element of [p, q, r] is $r$, which is equal to $c$. However, the third element of [c, b, a] is $a$. Therefore, option D is also not correct.

In conclusion, the correct answer is $\boxed{\text{(C)}}$.