Let a=f′(1), b=f′′(2), c=f′′′(3). Then f(x)=x3+ax2+2bx+c.
Taking derivatives: f′(x)=3x2+2ax+2b, f′′(x)=6x+2a, f′′′(x)=6.
From f′(1)=a: 3+2a+2b=a gives a=−3−2b.
From f′′(2)=b: 12+2a=b.
From f′′′(3)=c: c=6.
Substituting the first into the second: b=12+2(−3−2b)=6−4b gives 5b=6, so b=56.
Then a=−3−512=−527.
f(x)=x3−527x2+512x+6
f′(x)=3x2−554x+512
f′(5)=75−54+512=5117