Observe that, the given line # x=y=0" (i.e., the Z-"# Axis ) and
the plane # 2x+3y+4z-12=0# intersects each other in the
point #C=C(0,0,3)#.
Clearly, #C# is one vertex of the tetrahedron in question.
Similarly, #A=A(6,0,0), B=B(0,4,0) and O=O(0,0,0)# are
the other vertices.
So, if #P(x,y,z) and r# are respectively the centre and the radius
of the sphere #S# circumscribing the tetrahedron #OABC#, then,
#OP^2=OA^2=OB^2=OC^2=r^2#.
#:. x^2+y^2+z^2=(x-6)^2+y^2+z^2=x^2+(y-4)^2+z^2=x^2+y^2+(z-3)^2#.
Now, #OP^2=OA^2 rArr x^2=(x-6)^2 rArr x=3#.
Likewise, #y=2, z=3/2," giving the centre "P=P(3,2,3/2)#.
#:. r^2=OP^2=3^2+2^2+(3/2)^2=61/4#.
#"Therefore, "S : (x-3)^2+(y-2)^2+(z-3/2)^2=61/4, or, #
# S : x^2+y^2+z^2-6x-4y-3z=0#.
Enjoy Maths.!
