The correct answer is $\boxed{5 + j}$.
To find the eigenvalues of a matrix, we can use the characteristic equation. The characteristic equation for a 3×3 matrix is given by:
$$|A – \lambda I| = 0$$
where $A$ is the matrix and $\lambda$ is the eigenvalue.
In this case, we have:
$$\begin{vmatrix} {10}&{5 + {\text{j}}}&4 \ {\text{x}}&{20}&2 \ 4&2&{ – 10} \end{vmatrix} – \lambda \begin{vmatrix} {1}&{0}&{0} \ {0}&{1}&{0} \ {0}&{0}&{1} \end{vmatrix} = 0$$
Expanding the determinant, we get:
$$-\lambda^3 + 35 \lambda^2 – 250 \lambda + 1200 = 0$$
Solving for $\lambda$, we get the following eigenvalues:
$$\lambda = 5 + j, \lambda = 5 – j, \lambda = -10$$
Since the eigenvalues are all real, the value of $x$ for which all the eigenvalues of the matrix are real is $\boxed{5 + j}$.
Here is a brief explanation of each option:
- Option A: $5 + j$ is an eigenvalue of the matrix.
- Option B: $5 – j$ is an eigenvalue of the matrix.
- Option C: $1 – 5j$ is not an eigenvalue of the matrix.
- Option D: $1 + 5j$ is not an eigenvalue of the matrix.