(a) The gradient of \(PQ\) is

\( \frac{6-p}{8-0}=\frac{6-p}{8} \).

The gradient of \(QR\) is

\( \frac{10-6}{r-8}=\frac4{r-8} \).

Since \(PQ\) is perpendicular to \(QR\), the product of the gradients is \(-1\):

\( \frac{6-p}{8}\cdot\frac4{r-8}=-1 \).

\( 6-p=-2(r-8) \).

\( 6-p=-2r+16 \).

Therefore \( p=2r-10 \).

(b) Since \(PR=\sqrt{85}\),

\( r^2+(10-p)^2=85 \).

Use \(p=2r-10\):

\( r^2+(10-(2r-10))^2=85 \).

\( r^2+(20-2r)^2=85 \).

\( r^2+4(r-10)^2=85 \).

\( 5r^2-80r+315=0 \).

Divide by \(5\):

\( r^2-16r+63=0 \).

\( (r-7)(r-9)=0 \).

So \( r=7 \) or \( r=9 \).

If \( r=7 \), then \( p=2(7)-10=4 \).

If \( r=9 \), then \( p=2(9)-10=8 \).

Answer: \( (p,r)=(4,7) \) or \( (p,r)=(8,9) \).