By the very definition of the dirac distribution, we have for any function $f \in C^0(\mathbf R^3)$ that $$ \int_{\mathbf R^3} f(x)\delta(x-a)\, dx = f(a) $$ In your case, we have $$ f(x) = \def\abs#1{\left|#1\right|}\abs x^2 + \langle x,a\rangle + \abs a^2 $$ Hence, \begin{align*} \int_{\mathbf R^3}\bigl(\abs x^2 + \langle x,a\rangle + \abs a^2\bigr) \delta(x-a)\, dx &= f(a)\\\ &= \abs a^2 +\langle a,a\rangle + \abs a^2\\\ &= 3\abs a^2 \end{align*} So, you are correct. \end{align*}