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_(2)+l_(2)+l_(3))=(y)/(m_(1)+m_(2)+m_(3))=(z)/(n_(1)+n_(2)+n_(3))`.

Explanation

If `theta` is the `angleBCO`, then the direction consines of OA'
`("bisector of "angleBOC)" " are" "(l_(2)+l_(3))/(2cos(theta//2)),(m_(2)+m_(3))/(2cos(theta//2))`
`(n_(2)+n_(3))/(2cos(theta//2))` or the direction ratios of OA' are `l_(2)+l_(3),m_(2)+m_(3)andn_(2)+n_(3)`
Also, the direction cosines of OA are `l_(1),m_(1)andn_(1)`. Hence, the equation of plane AOA' is
`|{:(x,y,z),(l_(2)+l_(3),m_(2)+m_(3),n_(2)+n_(3)),(l_(1),m_(1),n_(1)):}|=0`
Applying `R_(2)toR_(2)+R_(3)`, we get the equation of plane AOA' as
`|{:(x,y,z),(l_(1)+l_(2)+l_(3),m_(1)+m_(2)+m_(3),n_(1)+n_(2)+n_(3)),(l_(1),m_(1),n_(1)):}|=0`
For all values of r, the point `(l_(1)+l_(2)+l_(3))r,(m_(1)+m_(2)+m_(3))and(n_(1)+n_(2)+n_(3))r)` lies on plane AOA'. Hence, the line `(x)/(l_(1)+l_(2)+l_(3))=(y)/(m_(1)+m_(2)+m_(3))=(z)/(n_(1)+n_(2)+n_(3))=r`
lies on plane AOA'. Similarly, this line lies on planes BOB' and COC' also. Hence, all the three planes, AOA' BOB' and COC', pass through the line.