Let position vectors of point A,B and C of triangle ABC represents be `hati+hatj+2hatk, hati+2hatj+hatk` and `2hati+hatj+hatk`. Let `l_(1)+l_(2)` and `l_(3)` be the length of perpendicular drawn from the orthocenter 'O' on the sides AB, BC and CA, then `(l_(1)+l_(2)+l_(3))` equals

Options

A. `2/sqrt(6)`
B. `3/sqrt(6)`
C. `sqrt(6)/2`
D. `sqrt(6)/3`

Answer

C

Explanation

Clearly triangle formed by the given points
`hati+hatj+2hatk, hati+2hatj+hatk` and `2hati+hatj+hatk` is equilateral as AB=BC=AC`=sqrt(2)`
`therefore` Distance of orthocenter 'O' from the sides is equal to inradius of the triangle.
`therefore l_(1)=l_(2)=l_(3)`= inradius `=r=Delta/s=(sqrt(3)/4(sqrt(2))^(2)/(3/2(sqrt(2))))=1/sqrt(6)`
`rArr (I_(1)+I_(2)+I_(3))=3/sqrt(6)=sqrt(6)/2`