O is the origin and lines OA, OB and OC have direction cosines `l_(r),m_(r)andn_(r)(r=1,2 and3)`. If lines OA', OB' and OC' bisect angles BOC, COA and AOB, respectively, prove that planes AOA', BOB' and COC' pass through the line `(x)/(l_(1)+l_(2)+l_(3))=(y)/(m_(1)+m_(2)+m_(3))=(z)/(n_(1)+n_(2)+n_(3))`.

Explanation

To solve the problem, we need to prove that the planes AOA', BOB', and COC' pass through the line defined by the equations \( \frac{x}{l_1 + l_2 + l_3} = \frac{y}{m_1 + m_2 + m_3} = \frac{z}{n_1 + n_2 + n_3} \).

### Step 1: Understanding the Direction Cosines
Given the direction cosines of lines OA, OB, and OC as \( l_r, m_r, n_r \) for \( r = 1, 2, 3 \), we can denote:
- OA: \( (l_1, m_1, n_1) \)
- OB: \( (l_2, m_2, n_2) \)
- OC: \( (l_3, m_3, n_3) \)

...