\[\boxed{\mathbf{187.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[По\ теореме\ о\ диагонали\ \]
\[прямоугольного\ \]
\[параллелепипеда:\]
\[d = \sqrt{a^{2} + b^{2} + c^{2}}.\]
\[\textbf{а)}\ a = 1;b = 1;c = 2:\]
\[d = \sqrt{1^{2} + 1^{2} + 2^{2}} = \sqrt{6}.\]
\[\textbf{б)}\ a = 8;b = 9;c = 12:\]
\[d = \sqrt{8^{2} + 9^{2} + 12^{2}} =\]
\[= \sqrt{64 + 81 + 144} = \sqrt{289} = 17.\]
\[\textbf{в)}\ a = \sqrt{39};\ \ b = 7;\ \ c = 9:\]
\[d = \sqrt{\left( \sqrt{39} \right)^{2} + 7^{2} + 9^{2}} =\]
\[= \sqrt{39 + 49 + 81} = \sqrt{169} = 13.\ \]