$f\left(x+y\right)=f\left(x\right)+3y^2+kxy$..(i)

In (i), put $x=1,f\left(1+y\right)=f\left(1\right)+3y^2+ky$

$\Rightarrow f\left(y+1\right)=3y^2+ky+1$.......(ii)

In (ii), put $y=1,f\left(2\right)=3\left(1\right)+k+1=6$

$\Rightarrow k=2$

Using (ii), $f\left(y+1\right)=3y^2+2y+1$

Replace $y\to x-1$

$\Rightarrow f\left(x\right)=3{\left(x-1\right)}^2+2\left(x-1\right)+1=3{\left(x-1\right)}^2+2x-1$

Now ${\int }_{-1}^{2}f\left(x\right)dx={\int }_{-1}^2\left(3{\left(x-1\right)}^2+2x-1\right)dx$$={\left[{\left(x-1\right)}^3+x^2-x\right]}_{-1}^2=\left[\left(8+1\right)+\left(4-1\right)-\left(2+1\right)\right]=13$