#### Answer

$\sqrt{6}-\sqrt{14}+\sqrt{21}-7$

#### Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$, or the product of 2 binomials, and the properties of radicals, the given expression, $
(\sqrt{2}+\sqrt{7})\sqrt{3}-\sqrt{7})
,$ is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{2}(\sqrt{3})+\sqrt{2}(-\sqrt{7})+\sqrt{7}(\sqrt{3})+\sqrt{7}(-\sqrt{7})
\\\\=
\sqrt{2(3)}-\sqrt{2(7)}+\sqrt{7(3)}-\sqrt{7(7)}
\\\\=
\sqrt{6}-\sqrt{14}+\sqrt{21}-\sqrt{(7)^2}
\\\\=
\sqrt{6}-\sqrt{14}+\sqrt{21}-7
.\end{array}