Let S≡(x1,y1)
Given:
P≡(−3,2)Q≡(9,10)R≡(α,4)

Since, PQ⊥RQ, so
mPQ×mQR=−1
⇒(9+310−2)×(9−α10−4)=−1
⇒(9−α1)=−41
⇒α=13
So, R≡(13,4)
So, centre of circle is O≡(213−3,24+2)≡(5,3)
Also,
mOQ⋅mQS=−1
⇒(5−93−10)mQS=−1
⇒mQS=−74
Equation of QS is
y−10=−74(x−9)
⇒4x+7y=106…(1)
Also,
mOR⋅mRS=−1
⇒(13−54−3)mRS=−1
⇒mRS=−8
Equation of RS
y−4=−8(x−13)
⇒8x+y=108…(2)
Solving equations (1)&(2), we get
x1=225;y1=8
Since, S(x1,y1) lies on 2x−ky=1
25−8k=1
⇒8k=24
⇒k=3