$OA,OB,OC$ are mutually perpendicular lines through the origin and their direction cosines are $l_1,m_1,n_1;l_2,m_2,n_2;l_3,m_3,n_3.$If $OA=a,OB=b,OC=c,$prove that the equation of the sphere $OABC$ is $x^2+y^2+z^2-x(al_1+bl_2+cl_3)-y(am_1+bm_2+cm_3)-z(an_1+bn_2+cn_3)=0$
The coordinates of $O(0,0,0),A(al_1,am_1,an_1),B(bl_2,bm_2,bn_2),C(cl_3,cm_3,cn_3)$.Let the equation of the sphere be $x^2+y^2+z^2+2ux+2vy+2wz+d=0$
Putting $O(0,0,0)$,we get $d=0$
Putting $A(al_1,am_1,an_1)$ in the equation of sphere,
$a^2l_1^2+a^2m_1^2+a^2n_1^2+2ual_1+2vam_1+2wan_1=0$
Putting $B(bl_2,bm_2,bn_2)$ in the equation of sphere,
$b^2l_2^2+b^2m_2^2+b^2n_2^2+2ubl_2+2vbm_2+2wbn_2=0$
Putting $A(cl_3,cm_3,cn_3)$ in the equation of sphere,
$c^2l_3^2+c^2m_3^2+c^2n_3^2+2ucl_3+2vcm_3+2wcn_3=0$
I am stuck now.Is my method correct.Is there less complicated method possible?Please help.
Consider the cuboid formed by the neighbouring vertices $O,A,B,C$.
Now consider the rectangle $OADB$ where $D$ is one of the vertices of the cuboid.
Since $AB$ is a diameter of the circle through $O, A, B$, $D$ must also lie on this circle and hence on the sphere.
By similar argument, all eight vertices of the cuboid must lie on the sphere, in particular the vertex at the opposite corner to $O$, which has coordinates $$ \underline{a}+\underline{b}+\underline{c}$$ Therefore the centre of the sphere has coordinates $$\frac 12\left(\underline{a}+\underline{b}+\underline{c}\right)$$
The required equation of the sphere then follows immediately.