Determinant Calculation

First, we evaluate the determinant of the given matrix:

Let \(D = \begin{vmatrix} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{vmatrix}\). Expanding the determinant, we get: \(D = x(yz - 1 \times 1) - 1(1 \times z - 1 \times 1) + 1(1 \times 1 - y \times 1)\) \(D = x(yz - 1) - (z - 1) + (1 - y)\) \(D = xyz - x - z + 1 + 1 - y\) \(D = xyz - x - y - z + 2\)

Applying Given Conditions

We are given that the determinant \(D\) is positive:

\(D > 0\)

Substituting the calculated expression for \(D\):

\(xyz - x - y - z + 2 > 0\)

Relating to p and q

We are also given the definitions \(p = x + y + z\) and \(q = xyz\). Substitute these into the inequality:

\(q - (x + y + z) + 2 > 0\)

\(q - p + 2 > 0\)

Final Inequality

Rearranging the terms to match the options:

\(q + 2 > p\)

This inequality matches Option C.