Calculating Direction Cosines of the Sides of a Triangle
To find the direction cosines of the sides of the triangle formed by the given vertices, we first need to calculate the direction vectors of each side. The direction vector of a line passing through two points A(x1, y1, z1) and B(x2, y2, z2) is given by the vector AB = (x2 - x1)i + (y2 - y1)j + (z2 - z1)k.

Calculating Direction Vector of Side 1
Given vertices: A(3, 5, -4) and B(-1, 1, 2)
Direction vector AB = (-1 - 3)i + (1 - 5)j + (2 - (-4))k
= -4i - 4j + 6k

Calculating Direction Vector of Side 2
Given vertices: B(-1, 1, 2) and C(-5, -5, -2)
Direction vector BC = (-5 + 1)i + (-5 - 1)j + (-2 - 2)k
= -4i - 6j - 4k

Calculating Direction Vector of Side 3
Given vertices: C(-5, -5, -2) and A(3, 5, -4)
Direction vector CA = (3 + 5)i + (5 + 5)j + (-4 - (-2))k
= 8i + 10j - 2k

Calculating Direction Cosines
The direction cosines of a vector are the cosines of the angles that the vector makes with the positive x, y, and z axes.
For a vector v = ai + bj + ck, the direction cosines are l = a/|v|, m = b/|v|, and n = c/|v|.

Direction Cosines of Side 1
|AB| = sqrt((-4)^2 + (-4)^2 + 6^2) = sqrt(16 + 16 + 36) = sqrt(68)
Direction cosines: l = -4/sqrt(68), m = -4/sqrt(68), n = 6/sqrt(68)

Direction Cosines of Side 2
|BC| = sqrt((-4)^2 + (-6)^2 + (-4)^2) = sqrt(16 + 36 + 16) = sqrt(68)
Direction cosines: l = -4/sqrt(68), m = -6/sqrt(68), n = -4/sqrt(68)

Direction Cosines of Side 3
|CA| = sqrt(8^2 + 10^2 + (-2)^2) = sqrt(64 + 100 + 4) = sqrt(168)
Direction cosines: l = 8/sqrt(168), m = 10/sqrt(168), n = -2/sqrt(168)
Therefore, the direction cosines of the sides of the triangle are as calculated above.