Show that the points A(2,-1,3),B(1,-3,1) and C(0,1,2) are the vertices of an isosceles right angled triangle.

Explanation

To show that the points A(2, -1, 3), B(1, -3, 1), and C(0, 1, 2) are the vertices of an isosceles right-angled triangle, we will follow these steps:

### Step 1: Calculate the lengths of the sides AB, BC, and AC using the distance formula.

The distance formula in three-dimensional space is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

#### Length of AB:
Let \( A(2, -1, 3) \) and \( B(1, -3, 1) \).

\[ AB = \sqrt{(1 - 2)^2 + (-3 + 1)^2 + (1 - 3)^2} \]
\[ = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} \]
\[ = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

#### Length of BC:
Let \( B(1, -3, 1) \) and \( C(0, 1, 2) \).

\[ BC = \sqrt{(0 - 1)^2 + (1 + 3)^2 + (2 - 1)^2} \]
\[ = \sqrt{(-1)^2 + (4)^2 + (1)^2} \]
\[ = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2} \]

#### Length of AC:
Let \( A(2, -1, 3) \) and \( C(0, 1, 2) \).

\[ AC = \sqrt{(0 - 2)^2 + (1 + 1)^2 + (2 - 3)^2} \]
\[ = \sqrt{(-2)^2 + (2)^2 + (-1)^2} \]
\[ = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 2: Compare the lengths of the sides.

We have:
- \( AB = 3 \)
- \( BC = 3\sqrt{2} \)
- \( AC = 3 \)

### Step 3: Check for isosceles condition.

Since \( AB = AC \), we have two sides equal, which satisfies the isosceles triangle condition.

### Step 4: Check for right angle using Pythagorean theorem.

For a right-angled triangle, the square of the length of the hypotenuse (the longest side) should equal the sum of the squares of the other two sides.

Here, the longest side is \( BC \):
\[ BC^2 = (3\sqrt{2})^2 = 18 \]
\[ AB^2 + AC^2 = 3^2 + 3^2 = 9 + 9 = 18 \]

Since \( AB^2 + AC^2 = BC^2 \), it confirms that the triangle is right-angled at point A.

### Conclusion:
The points A(2, -1, 3), B(1, -3, 1), and C(0, 1, 2) form an isosceles right-angled triangle.

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