If (6,10,10), (1,0,-5), (6,-10,0) are vertices of a triangle, find the direction ratios of its sides. Determine whether it is right-angled or isosceles.

We are given the vertices of a triangle:
A(6, 10, 10), B(1, 0, -5), and C(6, -10, 0).

Step 1: Find the direction ratios of the sides
Direction ratios of a line joining two points can be found by subtracting the coordinates of one point from the coordinates of the other. The direction ratios of a line joining points P(x1, y1, z1) and Q(x2, y2, z2) are given by:

Direction ratios = (x2 - x1, y2 - y1, z2 - z1)

Direction ratios of side AB:
Let the points be A(6, 10, 10) and B(1, 0, -5).

Direction ratios of AB = (1 - 6, 0 - 10, -5 - 10) = (-5, -10, -15)

Direction ratios of side BC:
Let the points be B(1, 0, -5) and C(6, -10, 0).

Direction ratios of BC = (6 - 1, -10 - 0, 0 - (-5)) = (5, -10, 5)

Direction ratios of side CA:
Let the points be C(6, -10, 0) and A(6, 10, 10).

Direction ratios of CA = (6 - 6, 10 - (-10), 10 - 0) = (0, 20, 10)

Step 2: Check if the triangle is right-angled or isosceles
To check if the triangle is right-angled, we can use the dot product of the direction ratios of two sides. If the dot product is zero, the two sides are perpendicular, and the triangle is right-angled.

Check if AB and BC are perpendicular:
Dot product of AB and BC = (-5 * 5) + (-10 * -10) + (-15 * 5) = -25 + 100 - 75 = 0

Since the dot product is zero, sides AB and BC are perpendicular, which means the triangle is right-angled.

Check if the triangle is isosceles:
For the triangle to be isosceles, at least two sides must have the same length.

Length of AB = √((-5)^2 + (-10)^2 + (-15)^2) = √(25 + 100 + 225) = √350

Length of BC = √(5^2 + (-10)^2 + 5^2) = √(25 + 100 + 25) = √150

Length of CA = √(0^2 + 20^2 + 10^2) = √(0 + 400 + 100) = √500

Since all three sides have different lengths, the triangle is not isosceles.

Final Answer:
The direction ratios of the sides of the triangle are:
AB: (-5, -10, -15)
BC: (5, -10, 5)
CA: (0, 20, 10)
The triangle is right-angled but not isosceles.