As #S and T# are points of tangency of the two circles, #PS and RT# are perpendicular to #AC, => PR# // #AC, => ST=PR=2r#
#DeltaPQR and DeltaABC# are similar,
#=> (QR)/(PR)=(BC)/(AC)#,
#=> QR=2r*3/5=(6r)/5#
Similarly, #(PQ)/(PR)=(AB)/(AC)#,
#=> PQ=2r*4/5=(8r)/5#
Now, #AS+ST+TC=AC, => x+2r+y=5#,
#=> color(red)(x=5-2r-y)#
SImilarly, #UP+QR+VC=BC, => r+(6r)/5+y=3#,
#=> color(red)((11r)/5+y=3 ----- Eq(1))#
Similarly, #AU+PQ+RV=AB, => x+(8r)/5+r=4#,
#=> (5-2r-y)+(8r)/5+r=4#
#=> color(red)((3r)/5-y=-1 ----- Eq(2))#
#Eq(1)+Eq(2), => (14r)/5=2#
#=> r=(2xx5)/14=5/7# units