The two eigenvalues of the matrix $\left[ {\begin{array}{*{20}{c}} 2&1 \ 1&{\text{p}} \end{array}} \right]$ are given by the solutions to the equation $|{A – \lambda }{I}| = 0$, where $A$ is the matrix and $\lambda$ is the eigenvalue. Expanding the determinant, we get:
$$\begin{align}
|{A – \lambda }{I}| &= \left| {\begin{array}{{20}{c}} 2&1 \ 1&{\text{p}} \end{array}} – \lambda \begin{array}{{20}{c}} 1&0 \ 0&1 \end{array}} \right| \\
&= \left| {\begin{array}{{20}{c}} 2-\lambda &1 \ 1&{\text{p}} – \lambda \end{array}} \right| \\
&= (\lambda – 2)(\text{p} – \lambda ) – 1 \\
&= \lambda ^2 – 3\lambda + 2{\text{p}} – 1.
\end{align*}$$
For the eigenvalues to have a ratio of 3 : 1, we need the discriminant of the quadratic equation $\lambda ^2 – 3\lambda + 2{\text{p}} – 1 = 0$ to be equal to 0. This gives us:
$$(-3)^2 – 4 \times 1 \times (2{\text{p}} – 1) = 0.$$
Solving for $\text{p}$, we get:
$$\text{p} = \frac{7}{3}.$$
Therefore, another value of $\text{p}$ for which the eigenvalues have the same ratio of 3 : 1 is $\frac{7}{3}$.
The other options are incorrect because they do not satisfy the condition that the eigenvalues have a ratio of 3 : 1.