We are given that `f''(x)=6(x-1)` , that the function has the point (2,1), and that the function is tangent to the line 3x-y-5=0 at (2,1).

So `f''(x)=6x-6`

Then `f'(x)=3x^2-6x+k` by taking an antiderivative.

Now the function is tangent to the line y=3x-5, so the slope of the tangent line is 3 at the point (2,1). Thus `f'(2)=3` .

`3(2)^2-6(2)+k=3==>k=3`

Now `f'(x)=3x^2-6x+3`

So `f(x)=x^3-3x^2+3x+k`

Since (2,1) lies on the curve, f(2)=1.

`(2)^3-3(2)^2+3(2)+k=1 ==>k=-1`

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**The equation of the curve is
`f(x)=x^3-3x^2+3x-1`**

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The graph of the curve and the tangent line (in red):

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