We are given the position vectors \( \mathbf{OA = i + 2j - 2k} \) and \( \mathbf{OB = -2i - 3j + 6k} \).
We need to find the position vector of point C, which lies on the bisector of \( \angle AOB \).
Step 1: Find the direction ratios of vectors \( \mathbf{OA} \) and \( \mathbf{OB} \)
The direction ratios of the vectors \( \mathbf{OA} \) and \( \mathbf{OB} \) are the coefficients of \( i, j, k \) in their respective expressions.
For \( \mathbf{OA} = i + 2j - 2k \), the direction ratios are \( (1, 2, -2) \).
For \( \mathbf{OB} = -2i - 3j + 6k \), the direction ratios are \( (-2, -3, 6) \).
Step 2: Use the formula for the bisector
The position vector \( \mathbf{OC} \) is given by the formula for the bisector of \( \angle AOB \):
\[
\mathbf{OC} = \frac{\mathbf{OA}}{\|\mathbf{OA}\|} + \frac{\mathbf{OB}}{\|\mathbf{OB}\|}
\]
First, calculate the magnitudes of \( \mathbf{OA} \) and \( \mathbf{OB} \):
\[
\|\mathbf{OA}\| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3
\]
\[
\|\mathbf{OB}\| = \sqrt{(-2)^2 + (-3)^2 + 6^2} = \sqrt{49} = 7
\]
Step 3: Find the vector \( \mathbf{OC} \)
Now substitute the values into the formula for \( \mathbf{OC} \):
\[
\mathbf{OC} = \frac{1}{3}(i + 2j - 2k) + \frac{1}{7}(-2i - 3j + 6k)
\]
Simplify the terms:
\[
\mathbf{OC} = \left( \frac{1}{3}i + \frac{2}{3}j - \frac{2}{3}k \right) + \left( -\frac{2}{7}i - \frac{3}{7}j + \frac{6}{7}k \right)
\]
Combine the components:
\[
\mathbf{OC} = \left( \frac{1}{3} - \frac{2}{7} \right)i + \left( \frac{2}{3} - \frac{3}{7} \right)j + \left( -\frac{2}{3} + \frac{6}{7} \right)k
\]
Step 4: Simplify the expression
First, calculate the coefficients:
\[
\frac{1}{3} - \frac{2}{7} = \frac{7 - 6}{21} = \frac{1}{21}
\]
\[
\frac{2}{3} - \frac{3}{7} = \frac{14 - 9}{21} = \frac{5}{21}
\]
\[
-\frac{2}{3} + \frac{6}{7} = \frac{-14 + 18}{21} = \frac{4}{21}
\]
Thus, \( \mathbf{OC} = \frac{1}{21}i + \frac{5}{21}j + \frac{4}{21}k \).
Multiply by 21 to remove the denominator:
\[
\mathbf{OC} = i + 5j + 4k
\]
Thus, the correct answer is option (2).