Find the values of a such that PQ = QR, where P, Q and R are the points whose coordinates are (6, - 1), (1, 3), and (a, 8).

Given Points: P(6, - 1), Q(1, 3) and R(a, 8).

We know distance formula to find the distance between two points is 

d =  $\sqrt{({\mathrm{x}}_2-{\mathrm{x}}_1{)}^{2 }+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2 }$

Where $(x_1, y_1) and (x_2,y_2)$coordinates of two points.

Now find distance between point P and Q.

$\mathrm{PQ}=\sqrt{{\left(6-1\right)}^2+{\left(-1-3\right)}^2}$

$\mathrm{PQ}=\sqrt{{\left(5\right)}^2+{\left(-4\right)}^2}$

$=\sqrt{25+16}$

$=\sqrt{41} \mathrm{unit}$s

Now find distance between point Q and R.

$\mathrm{QR}=\sqrt{{\left(1-\mathrm{a}\right)}^2+{\left(3-8\right)}^2}$

$=\sqrt{{\left(1-\mathrm{a}\right)}^2+{\left(5\right)}^2}$

$=\sqrt{{\left(1-\mathrm{a}\right)}^2+25}$

Given, PQ = QR

$\Rightarrow \sqrt{41}=\sqrt{{\left(1-\mathrm{a}\right)}^2+25}$

Squaring on both side, we get:

$\Rightarrow 41={\left(1-\mathrm{a}\right)}^2+25$

$\Rightarrow {\left(1-\mathrm{a}\right)}^2=41-25$

$\Rightarrow {\left(1-\mathrm{a}\right)}^2=16$

$\Rightarrow \left(1-\mathrm{a}\right)=\pm 4$

By Comparing both sides, we get:

$\Rightarrow \left(1-\mathrm{a}\right)= 4, \left(1-\mathrm{a}\right)=-4$

$\Rightarrow \mathrm{a}= -3, \mathrm{a}=5$