Easy way to do the Laplacian here

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Let $\mathbf p$ be a constant vector.

Let $\mathbf r = (x,y,z) $ be the position vector

Let $r^2 = x^2+y^2+z^2$.

How to compute $\nabla^2 \bigg(\dfrac{\mathbf p \cdot \mathbf r}{r}\bigg)$?

Obviously, one way is to stick with Cartesian coordinates and just bash it out, but I'm looking for a way to use spherical polars and vector identities?

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$\mathbf p\cdot\mathbf r=pr\cos\theta$, where $\theta$ is the angle between $\mathbf p$ and $\mathbf r$. You can use a coordinate system where $\mathbf p$ points in the $z$ direction ($\theta=0$). Then all you need is $\nabla^2(\cos\theta)$. If you want a shortcut, you can find an expression for the Laplacian in spherical coordinates on the web.