If three mutually perpendicular lines have direction cosines `(l_1,m_1,n_1),(l_2,m_2,n_2) and (L_3 ,m_3, n_3)`, then the line having direction cosines `l_1+l_2+l_3,m_1+ m_2+m_3, and n_1 + n_2 + n_3`, make an angle of

Explanation

According to the question, OA, OB and OC are mutually perpendicular to each other,
hence `l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=0`
`l_(2)l_(3)+m_(2)m_(3)+n_(2)n_(3)= 0`
`l_(3)l_(1)+m_(3)m_(1)+n_(3)n_(1)= 0`
Let l, m, n be the direction cosines of OP, then
`1/(l_(1)+l_(2)+l_(3))=m/(m_(1)+m_(2)+m_(3))=n/(n_(1)+n_(2)+n_(3))`
`=(sqrt(l^(2)+m^(2)+n^(3)))/(sqrt(l_(1)^(2)+m_(1)^(2)+n_(1)^(2)+l_(2)^(2)+m_(2)^(2)+n_(2)^(2)+l_(3)^(3)+m_(3)^(3)+n_(3)^(3)))=1/sqrt(3)`
`rArr l=(l_(1)+l_(2)+l_(3))/sqrt(3),m=(m_(1)+m_(2)+m_(3))/sqrt(3),n=(n_(1)+n_(2)+n_(3))/sqrt(3)`
Let `theta`, be the angle between OA and OP, then
`cos theta = ll_(1)+mm_(1)+n n_(1)=(l_(1)+l_(2)+l_(3))/sqrt(3)l_(1)+(m_(1)+m_(2)+m_(3))/sqrt(3)m_(1)+(n_(1)+n_(2)+n_(3))/sqrt(3)n_(1)=1/sqrt(3)`
Similarly, `phi and psi` be the angle between OB and OP, between OC and OP respectibely, then
`cos phi = ll_(2) +mm_(2)+n n_(2)=1/sqrt(3)`
`cos psi = ll_(3) +mm_(3)+n n_(3)=1/sqrt(3)`
`therefore cos theta = cos phi =cospsi`
`rArr ` OP makes equal angle with OA, OB, AND OC.