`A(1,-1,-3), B(2, 1,-2) & C(-5, 2,-6)` are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is :

Explanation

To find the length of the bisector of the internal angle at vertex A of triangle ABC with vertices A(1, -1, -3), B(2, 1, -2), and C(-5, 2, -6), we can follow these steps:

### Step 1: Find the vectors AB and AC
The position vectors of points A, B, and C are given as:
- A = (1, -1, -3)
- B = (2, 1, -2)
- C = (-5, 2, -6)

We can find the vectors AB and AC as follows:
\[ \vec{AB} = \vec{B} - \vec{A} = (2 - 1, 1 - (-1), -2 - (-3)) = (1, 2, 1) \]
\[ \vec{AC} = \vec{C} - \vec{A} = (-5 - 1, 2 - (-1), -6 - (-3)) = (-6, 3, -3) \]

### Step 2: Calculate the lengths of AB and AC
Now, we calculate the magnitudes of vectors AB and AC:
\[ |\vec{AB}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]
\[ |\vec{AC}| = \sqrt{(-6)^2 + 3^2 + (-3)^2} = \sqrt{36 + 9 + 9} = \sqrt{54} = 3\sqrt{6} \]

### Step 3: Use the angle bisector theorem
According to the angle bisector theorem, the length of the angle bisector AD from vertex A to side BC can be calculated using the formula:
\[ AD = \frac{2 \cdot |\vec{AB}| \cdot |\vec{AC}|}{|\vec{AB}| + |\vec{AC}|} \]

Substituting the lengths we found:
\[ AD = \frac{2 \cdot \sqrt{6} \cdot 3\sqrt{6}}{\sqrt{6} + 3\sqrt{6}} = \frac{2 \cdot 3 \cdot 6}{4\sqrt{6}} = \frac{36}{4\sqrt{6}} = \frac{9}{\sqrt{6}} = \frac{9\sqrt{6}}{6} = \frac{3\sqrt{6}}{2} \]

### Step 4: Final result
Thus, the length of the bisector of the internal angle at A is:
\[ AD = \frac{3\sqrt{6}}{2} \]