The path [tex]\mathcal C[/tex] can be parameterized by the vector-valued function,
[tex]\mathbf r(t)=t(2\,\mathbf i+3\,\mathbf j+9\,\mathbf k)[/tex]
where [tex]0\le t\le1[/tex]. Then
[tex]\mathrm d\mathbf r=(2\,\mathbf i+3\,\mathbf j+9\,\mathbf k)\,\mathrm dt[/tex]
The line integral reduces to
[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}(x(t)^3\,\mathbf i+y(t)^2\,\mathbf j+z(t)\,\mathbf k)\cdot(2\,\mathbf i+3\,\mathbf j+9\,\mathbf k)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_{t=0}^{t=1}((2t)^3\,\mathbf i+(3t)^2\,\mathbf j+9t\,\mathbf k)\cdot(2\,\mathbf i+3\,\mathbf j+9\,\mathbf k)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_{t=0}^{t=1}(16t^3+27t^2+81t)\,\mathrm dt=\frac{107}2[/tex]