Show that the points `A(3, -2,1), B(1,-3,5), C(2,1,-4)` do not form a right - angled triangle.

Answer

To show that the points \( A(3, -2, 1) \), \( B(1, -3, 5) \), and \( C(2, 1, -4) \) do not form a right-angled triangle, we will calculate the distances between each pair of points and check if the Pythagorean theorem holds.

### Step 1: Calculate the distance \( AB \)

Using the distance formula in 3D space:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

For points \( A(3, -2, 1) \) and \( B(1, -3, 5) \):
\[ AB = \sqrt{(1 - 3)^2 + (-3 + 2)^2 + (5 - 1)^2} \]
\[ = \sqrt{(-2)^2 + (-1)^2 + (4)^2} \]
\[ = \sqrt{4 + 1 + 16} = \sqrt{21} \]

### Step 2: Calculate the distance \( BC \)

For points \( B(1, -3, 5) \) and \( C(2, 1, -4) \):
\[ BC = \sqrt{(2 - 1)^2 + (1 + 3)^2 + (-4 - 5)^2} \]
\[ = \sqrt{(1)^2 + (4)^2 + (-9)^2} \]
\[ = \sqrt{1 + 16 + 81} = \sqrt{98} \]

### Step 3: Calculate the distance \( CA \)

For points \( C(2, 1, -4) \) and \( A(3, -2, 1) \):
\[ CA = \sqrt{(3 - 2)^2 + (-2 - 1)^2 + (1 + 4)^2} \]
\[ = \sqrt{(1)^2 + (-3)^2 + (5)^2} \]
\[ = \sqrt{1 + 9 + 25} = \sqrt{35} \]

### Step 4: Check the Pythagorean theorem

We need to check if any of the following conditions hold:
1. \( AB^2 + CA^2 = BC^2 \)
2. \( AB^2 + BC^2 = CA^2 \)
3. \( BC^2 + CA^2 = AB^2 \)

Calculating the squares:
- \( AB^2 = 21 \)
- \( BC^2 = 98 \)
- \( CA^2 = 35 \)

Now check the conditions:
1. \( AB^2 + CA^2 = 21 + 35 = 56 \) (not equal to \( 98 \))
2. \( AB^2 + BC^2 = 21 + 98 = 119 \) (not equal to \( 35 \))
3. \( BC^2 + CA^2 = 98 + 35 = 133 \) (not equal to \( 21 \))

Since none of these conditions hold, we conclude that the points \( A \), \( B \), and \( C \) do not form a right-angled triangle.

### Conclusion

The points \( A(3, -2, 1) \), \( B(1, -3, 5) \), and \( C(2, 1, -4) \) do not form a right-angled triangle.

---