Point P (-3, 2), Q (9, 10) and R (a, 4) lie on a circle C with PR as its diameter. The tangents to C at the
points Q and R intersect at the point S. If S lies on the line $2x-ky=1$ then $k$ is equal to

Equation of circle with PR as diameter is   $(x+3)(x-a)+(y-2)(y-4)=0$
$\mathrm{Q}(9,10) \mathrm{lies} \mathrm{on} \mathrm{the} \mathrm{circle}\Rightarrow a=13$
equation of circle $x^2+y^2-10x-6y-31=0$
Equation of tangent at $Q:4x+7y-106=0$
Equation of tangent R = $8x+y-108=0$…………(2) 
From (1), (2) (x, y) $=\left(\frac{25}{2},8\right)$
$\left(\frac{25}{2},8\right)\Rightarrow 2\mathrm{x}-\mathrm{ky}=1\Rightarrow 25-8\mathrm{k}=1\Rightarrow 8\mathrm{k}=24\Rightarrow \boxed{\mathrm{k}=3}$