The correct answer is $\boxed{\left[ {\begin{array}{{20}{c}} { – 6} \ { – 3} \ 6 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 4 \ { – 2} \ 3 \end{array}} \right]}$.
To find an orthogonal set of vectors having a span that contains P, Q, R, we can use the Gram-Schmidt process. The Gram-Schmidt process is a method for orthonormalizing a set of vectors. It works by iteratively projecting each vector onto the space spanned by the previous vectors, and then normalizing the resulting vector.
In this case, we start with the vectors P, Q, and R. We then project P onto the space spanned by Q and R. This gives us the vector $\mathbf{p}_1 = P – \frac{\langle P, Q \rangle}{\langle Q, Q \rangle} Q – \frac{\langle P, R \rangle}{\langle R, R \rangle} R$. We then normalize $\mathbf{p}_1$ to get the vector $\mathbf{v}_1$.
We then project Q onto the space spanned by $\mathbf{v}_1$ and R. This gives us the vector $\mathbf{q}_1 = Q – \frac{\langle Q, \mathbf{v}_1 \rangle}{\langle \mathbf{v}_1, \mathbf{v}_1 \rangle} \mathbf{v}_1 – \frac{\langle Q, R \rangle}{\langle R, R \rangle} R$. We then normalize $\mathbf{q}_1$ to get the vector $\mathbf{v}_2$.
We continue this process until we have projected all of the vectors onto the space spanned by the previous vectors. The resulting vectors are an orthogonal set of vectors having a span that contains P, Q, and R.
In this case, the resulting vectors are $\mathbf{v}_1 = \left[ {\begin{array}{{20}{c}} { – 6} \ { – 3} \ 6 \end{array}} \right]$ and $\mathbf{v}_2 = \left[ {\begin{array}{{20}{c}} 4 \ { – 2} \ 3 \end{array}} \right]$.
Therefore, the orthogonal set of vectors having a span that contains P, Q, and R is $\boxed{\left[ {\begin{array}{{20}{c}} { – 6} \ { – 3} \ 6 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 4 \ { – 2} \ 3 \end{array}} \right]}$.