Find the direction cosines of the sides of the triangle whose vertices are $(3,5,-4), (-1,1,2)$ and $(-5,-5,-2)$.

Let the vertices of $\triangle ABC$ be $A(3,5,-4), B(-1,1,2),$ and $C(-5,-5,-2)$.
The direction ratios of side $AB$ are $(-1-3), (1-5), (2-(-4)),$ which are $-4, -4, 6$.
The magnitude of vector $\vec{AB}$ is $\sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{16+16+36} = \sqrt{68} = 2\sqrt{17}$.
Thus,the direction cosines of $AB$ are $\frac{-4}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}},$ which simplify to $-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}}$.
The direction ratios of side $BC$ are $(-5-(-1)), (-5-1), (-2-2),$ which are $-4, -6, -4$.
The magnitude of vector $\vec{BC}$ is $\sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16+36+16} = \sqrt{68} = 2\sqrt{17}$.
Thus,the direction cosines of $BC$ are $\frac{-4}{2\sqrt{17}}, \frac{-6}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}},$ which simplify to $-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}}$.
The direction ratios of side $CA$ are $(3-(-5)), (5-(-5)), (-4-(-2)),$ which are $8, 10, -2$.
The magnitude of vector $\vec{CA}$ is $\sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{64+100+4} = \sqrt{168} = 2\sqrt{42}$.
Thus,the direction cosines of $CA$ are $\frac{8}{2\sqrt{42}}, \frac{10}{2\sqrt{42}}, \frac{-2}{2\sqrt{42}},$ which simplify to $\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}}$.