Compute the following integral over the surface of a three-dimensional sphere of unit radius: $$ I(k)=\int_{|r|=1} dS e^{ik*r}$$ Here dS denotes the area of a differentially small surface element $$ k=(k_{x}, k_{y}, k_{z})$$ is a given three-dimensional vector. the vector $$r=(x,y,z) $$ and the scalar product $$k*r=(x*k_{x}, x*k_{y}, x*k_{z})$$
For a sphere, it is easiest to invoke spherical coordinates to tackle a problem like this. In such a case, the area element on a surface of radius $r$ is given by $$dS=r^2 \sin\theta d\theta d\phi,$$ where $\theta$ runs from $0$ to $\pi$ and $\phi$ runs over $0$ to $2\pi$. This is also bound to the surface $r=1$ so that we only have $dS=\sin\theta d\theta d\phi$.
At this point I would note that your scalar product $\mathbf{k}\cdot\mathbf{r}$ should be $\textit{scalar}$ rather than a vector. If $\mathbf{r}=(x,y,z)$, then $$\mathbf{k}\cdot\mathbf{r}=k_xx+k_yy+k_zz.$$
Using spherical coordinates, this becomes $$\mathbf{k}\cdot\mathbf{r}=k_xr\sin\theta\cos\phi+k_yr\sin\theta\sin\phi+k_zr\cos\theta.$$
We can note that in the $\mathbf{k}$-dependence, there is rotational symmetry. You can see this from the definition of the scalar product, $\mathbf{k}\cdot\mathbf{r}=|\mathbf{k}||\mathbf{r}|\cos\theta$. So let's first rotate such that the $\mathbf{k}$ vector is pointing along the $z$-axis (such that $k_x=0$, $k_y=0$, $k_z=K=|\mathbf{k}$|).
Plugging this into the integral, and recalling that on the surface $r=|\mathbf{r}|=1$, we have $$ \begin{align} I(k)&=\int_0^{2\pi}\int_0^\pi e^{i(k_xr\sin\theta\cos\phi+k_yr\sin\theta\sin\phi+k_zr\cos\theta)}\sin\theta d\theta d\phi,\\ &=\int_0^{2\pi}\int_0^\pi e^{iK\cos\theta}\sin\theta d\theta d\phi. \end{align} $$ The integral over $\phi$ is immediately doable,
$$ I(k)=2\pi\int_0^\pi e^{iK\cos\theta}\sin\theta d\theta. $$ Then one should note that $$ \frac{d}{d\theta}e^{iK\cos\theta}=-i\sin\theta Ke^{iK\cos\theta}, $$ hence $$ \begin{align} I(k)&=2\pi\int_0^\pi e^{iK\cos\theta}\sin\theta d\theta,\\ &=-\frac{2\pi}{iK}\int_0^\pi \frac{d}{d\theta}\left(e^{iK\cos\theta}\right)d\theta,\\ &=-\frac{2\pi}{iK}\left(e^{iK\cos\pi}-e^{iK\cos 0}\right),\\ &=\frac{2\pi}{iK}\left(e^{iK}-e^{-iK}\right),\\ &=\frac{4\pi}{K}\frac{1}{2i}\left(e^{iK}-e^{-iK}\right),\\ &=\frac{4\pi}{K}\sin K,\\ &=4\pi\frac{\sin\left(\sqrt{k_x^2+k_y^2+k_z^2}\right)}{\sqrt{k_x^2+k_y^2+k_z^2}}, \end{align} $$ where in the above I have used the fundamental theorem of calculus and the definition of the sine function in terms of the exponential.