Step 1: Identifying Vertices and Defining Sides

Let the vertices of the triangle be denoted as follows:

• $A = (3, 5, -4)$
• $B = (-1, 1, 2)$
• $C = (-5, -5, -2)$

The sides of the triangle are the line segments $AB$, $BC$, and $CA$. The direction cosines of a line segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $\frac{x_2-x_1}{d}, \frac{y_2-y_1}{d}, \frac{z_2-z_1}{d}$, where $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ [Per the distance formula in 3D space].

Step 2: Calculating Direction Cosines for Side AB

Vector $\vec{AB} = (-1-3, 1-5, 2-(-4)) = (-4, -4, 6)$.

Length $d_{AB} = \sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{16 + 16 + 36} = \sqrt{68} = 2\sqrt{17}$.

Direction Cosines ($l_1, m_1, n_1$):

• $l_1 = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$
• $m_1 = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$
• $n_1 = \frac{6}{2\sqrt{17}} = \frac{3}{\sqrt{17}}$

Step 3: Calculating Direction Cosines for Side BC

Vector $\vec{BC} = (-5-(-1), -5-1, -2-2) = (-4, -6, -4)$.

Length $d_{BC} = \sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17}$.

Direction Cosines ($l_2, m_2, n_2$):

• $l_2 = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$
• $m_2 = \frac{-6}{2\sqrt{17}} = -\frac{3}{\sqrt{17}}$
• $n_2 = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$

Step 4: Calculating Direction Cosines for Side CA

Vector $\vec{CA} = (3-(-5), 5-(-5), -4-(-2)) = (8, 10, -2)$.

Length $d_{CA} = \sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{64 + 100 + 4} = \sqrt{168} = 2\sqrt{42}$.

Direction Cosines ($l_3, m_3, n_3$):

• $l_3 = \frac{8}{2\sqrt{42}} = \frac{4}{\sqrt{42}}$
• $m_3 = \frac{10}{2\sqrt{42}} = \frac{5}{\sqrt{42}}$
• $n_3 = \frac{-2}{2\sqrt{42}} = -\frac{1}{\sqrt{42}}$

Step 5: Summary Table of Results

Side	Direction Cosines $(l, m, n)$
$AB$	$(-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}})$
$BC$	$(-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}})$
$CA$	$(\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}})$

Final Solution: The direction cosines are as listed in the table above for sides AB, BC, and CA respectively.

Step 1: Theoretical Foundation of Direction Cosines

In three-dimensional coordinate geometry, if a line makes angles $\alpha$, $\beta$, and $\gamma$ with the positive directions of the $x$, $y$, and $z$-axes respectively, the direction cosines of the line are defined as:

• $l = \cos \alpha$
• $m = \cos \beta$
• $n = \cos \gamma$

[Definition of Direction Cosines in Cartesian Space]

Step 2: Substitution of Given Angles

Given the angles:

• $\alpha = 90^\circ$
• $\beta = 135^\circ$
• $\gamma = 45^\circ$

Step 3: Calculating Trigonometric Values

Applying the cosine function to each angle:

$l = \cos(90^\circ) = 0$

$m = \cos(135^\circ) = \cos(180^\circ - 45^\circ) = -\cos(45^\circ) = -\frac{1}{\sqrt{2}}$

$n = \cos(45^\circ) = \frac{1}{\sqrt{2}}$

[Trigonometric Identity: $\cos(180^\circ - \theta) = -\cos(\theta)$]

Step 4: Verification via Direction Cosine Property

The sum of the squares of the direction cosines must equal unity: $l^2 + m^2 + n^2 = 1$.

$0^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 0 + \frac{1}{2} + \frac{1}{2} = 1$

[Per the property of direction cosines: $l^2 + m^2 + n^2 = 1$]

Step 5: Visual Representation

Final Solution: The direction cosines of the line are $0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.

Step 1: Definition of Direction Ratios and Cosines

Given the direction ratios $(a, b, c)$ of a line as $-18, 12, -4$, the direction cosines $(l, m, n)$ are defined by the normalized components of the direction vector divided by its magnitude $r$.

The relations are defined as: $l = \frac{a}{r}, \quad m = \frac{b}{r}, \quad n = \frac{c}{r}$ where $r = \sqrt{a^2 + b^2 + c^2}$ [Per the definition of direction cosines in 3D Euclidean space].

Step 2: Calculation of the Magnitude ($r$)

Substitute the given values $a = -18, b = 12, c = -4$ into the magnitude formula:

$r = \sqrt{(-18)^2 + (12)^2 + (-4)^2}$

$r = \sqrt{324 + 144 + 16}$

$r = \sqrt{484}$

$r = 22$ [By the property of square roots where $\sqrt{484} = 22$].

Step 3: Determining the Direction Cosines

Since the direction cosines can be both positive and negative (depending on the direction of the line segment), we consider $r = \pm 22$.

For $r = 22$:

• $l = \frac{-18}{22} = -\frac{9}{11}$
• $m = \frac{12}{22} = \frac{6}{11}$
• $n = \frac{-4}{22} = -\frac{2}{11}$

For $r = -22$:

• $l = \frac{-18}{-22} = \frac{9}{11}$
• $m = \frac{12}{-22} = -\frac{6}{11}$
• $n = \frac{-4}{-22} = \frac{2}{11}$

Visual Representation of Direction Ratios

Verification via Identity

We verify using the identity $l^2 + m^2 + n^2 = 1$ [Fundamental property of direction cosines]:

$(-\frac{9}{11})^2 + (\frac{6}{11})^2 + (-\frac{2}{11})^2 = \frac{81}{121} + \frac{36}{121} + \frac{4}{121} = \frac{121}{121} = 1$.

Final Solution: The direction cosines are $\pm \left( -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right)$ or simply $\left( -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right)$ and $\left( \frac{9}{11}, -\frac{6}{11}, \frac{2}{11} \right)$.

Given Coordinates

Let the given points be denoted as:

• $A = (2, 3, 4)$
• $B = (-1, -2, 1)$
• $C = (5, 8, 7)$

Step 1: Theoretical Basis for Collinearity

Three points $A$, $B$, and $C$ are collinear if and only if the sum of the lengths of any two line segments formed by these points is equal to the length of the third segment (e.g., $AB + BC = AC$) [Per the Triangle Inequality Theorem]. Alternatively, we can demonstrate that the vectors $\vec{AB}$ and $\vec{BC}$ are scalar multiples of each other [Per the condition for parallel vectors].

Step 2: Calculating Distances using the Distance Formula

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Distance $AB$:

$AB = \sqrt{(-1 - 2)^2 + (-2 - 3)^2 + (1 - 4)^2}$ $AB = \sqrt{(-3)^2 + (-5)^2 + (-3)^2}$ $AB = \sqrt{9 + 25 + 9} = \sqrt{43}$

Distance $BC$:

$BC = \sqrt{(5 - (-1))^2 + (8 - (-2))^2 + (7 - 1)^2}$ $BC = \sqrt{(6)^2 + (10)^2 + (6)^2}$ $BC = \sqrt{36 + 100 + 36} = \sqrt{172} = \sqrt{4 \times 43} = 2\sqrt{43}$

Distance $AC$:

$AC = \sqrt{(5 - 2)^2 + (8 - 3)^2 + (7 - 4)^2}$ $AC = \sqrt{(3)^2 + (5)^2 + (3)^2}$ $AC = \sqrt{9 + 25 + 9} = \sqrt{43}$

Step 3: Verification of Collinearity

Observe the calculated magnitudes:

• $AB = \sqrt{43}$
• $BC = 2\sqrt{43}$
• $AC = \sqrt{43}$

Note that $AB + AC = \sqrt{43} + \sqrt{43} = 2\sqrt{43} = BC$.

Since $AB + AC = BC$, point $A$ lies on the line segment $BC$. Thus, the points are collinear.

Vector Representation (Visual Reference)

Final Solution: Since the sum of distances $AB$ and $AC$ equals the distance $BC$ ($\sqrt{43} + \sqrt{43} = 2\sqrt{43}$), the points $(2, 3, 4)$, $(-1, -2, 1)$, and $(5, 8, 7)$ are proven to be collinear.

Step 1: Establishing Geometric Variables

Let the line make an angle $\alpha$ with the $x$-axis, $y$-axis, and $z$-axis. Since the problem specifies that the line makes equal angles with the coordinate axes, we define the direction angles as follows:

• Angle with $x$-axis: $\alpha$
• Angle with $y$-axis: $\alpha$
• Angle with $z$-axis: $\alpha$

[Per the definition of direction angles in 3D Euclidean space]

Step 2: Defining Direction Cosines

The direction cosines $(l, m, n)$ of a line are defined as the cosines of the angles the line makes with the positive coordinate axes:

• $l = \cos \alpha$
• $m = \cos \alpha$
• $n = \cos \alpha$

Step 3: Applying the Fundamental Identity

For any line in three-dimensional space, the sum of the squares of the direction cosines is always equal to unity [By the property of direction cosines in $\mathbb{R}^3$]:

$l^2 + m^2 + n^2 = 1$

Step 4: Algebraic Substitution and Solving

Substitute the expressions from Step 2 into the identity:

$\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$

$3 \cos^2 \alpha = 1$

$\cos^2 \alpha = \frac{1}{3}$

Taking the square root on both sides:

$\cos \alpha = \pm \frac{1}{\sqrt{3}}$

Step 5: Visual Representation

Step 6: Determining the Set of Direction Cosines

Since $l = m = n = \cos \alpha$, we obtain two possible sets of direction cosines depending on the orientation of the line:

• Case 1: $l = \frac{1}{\sqrt{3}}, m = \frac{1}{\sqrt{3}}, n = \frac{1}{\sqrt{3}}$
• Case 2: $l = -\frac{1}{\sqrt{3}}, m = -\frac{1}{\sqrt{3}}, n = -\frac{1}{\sqrt{3}}$

Final Solution: The direction cosines of the line are $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$ or $\left( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right)$.