The centroid of a triangle is the average of the coordinates of its three vertices. Given the centroid $(1, 2, 3)$ and the vertices of the triangle $A(p, 5, 9)$, $B(4, q, -6)$, and $C(10, 8, r)$, we can use the formula for the coordinates of the centroid:

$\left( \frac{p + 4 + 10}{3}, \frac{5 + q + 8}{3}, \frac{9 - 6 + r}{3} \right) = (1, 2, 3)$

We now equate the coordinates to find $p$, $q$, and $r$.

Step 1: Find $p$, $q$, and $r$ using the centroid condition

From the $x$-coordinate: $\frac{p + 4 + 10}{3} = 1 \quad \Rightarrow \quad p + 14 = 3 \quad \Rightarrow \quad p = -11$

From the $y$-coordinate: $\frac{5 + q + 8}{3} = 2 \quad \Rightarrow \quad q + 13 = 6 \quad \Rightarrow \quad q = -7$

From the $z$-coordinate: $\frac{9 - 6 + r}{3} = 3 \quad \Rightarrow \quad r + 3 = 9 \quad \Rightarrow \quad r = 6$

Thus, we have $p = -11$, $q = -7$, and $r = 6$.

Step 2: Verify the points lie on the plane $2x - y + z = 7$

For vertex $A(-11, 5, 9)$: $2(-11) - 5 + 9 = -22 - 5 + 9 = -18 \quad (\text{not on the plane})$

For vertex $B(4, -7, -6)$: $2(4) - (-7) + (-6) = 8 + 7 - 6 = 9 \quad (\text{not on the plane})$

For vertex $C(10, 8, 6)$: [ 2(10) - 8 + 6 = 20 - 8 + 6 = 18 \quad (\text{not on the plane})