A variable plane is at a constant distance p from the origin and meets the coordinate axes in A,B and C , show that the locus of the centroid of the tetrahedron OABC os `1/x^2+1/y^2+1/z^2= 16/p^2`

A variable plane is at a constant distance p from the origin and meets the coordinate axes in A,B and C. Show that the locus of the centroid of the tetrahedron OABC is (1)/(x^2)+(1)/(y^2)+(1)/(z^2)=(16)/(p^2) .

A variable line is at constant distance p from the origin and meets the co-ordinate axes in A, B. Show that the locus of the centroid of the triangleOAB is x^(-2)+y^(-2)=9p^(-2)

A variable plane which is at a constant distance 3p from origin cuts the coordinate axes at A,B,C respectively. Show that locus of the centroid of the triangleABC is 1/x^2+1/y^2+1/z^2= 1/p^2 .

a variable plane which is at a constant distance 3p from the origin O cuts the axes at A,B,C. Show that the locus of the centroid of the triangle ABC is x^(-2)+y^(-2)+z^(-2)=p^(-2) .

A variable plane which is at a constant distance 3p from the origin O cuts the axes at L,M and N.Show that the locus of the points of intersection of the planes through L,M,N drwon parallel to the coordinate planes is 9(x^(-2)+y^(-2)+z^(-2))=p^(-2) .

A sphere of constant radius 2k passes through the origin and meets the axes in A ,B ,a n dCdot The locus of a centroid of the tetrahedron O A B C is a. x^2+y^2+z^2=4k^2 b. x^2+y^2+z^2=k^2 c. 2(x^2+y^2+z)^2=k^2 d. none of these

The tangent at any point P on the circle x^2+y^2=4 meets the coordinate axes at A and B . Then find the locus of the midpoint of A Bdot

A sphere of constant radius k passes through the origin and meets the axes at A, B and C. Prove that the centroid of triangle ABC lies on the sphere 9(x^(2)+y^(2)+z^(2))=4k^(2) .

A variable plane x/a+y/b+z/c=1 at a unit distance from origin cuts the coordinate axes at A, B and C. Centroid (x, y, z) satisfies the equation 1/x^2+1/y^2+1/z^2=K. The value of K is (A) 9 (B) 3 (C) 1/9 (D) 1/3

A circle of constant radius a passes through the origin O and cuts the axes of coordinates at points P and Q . Then the equation of the locus of the foot of perpendicular from O to P Q is (A) (x^2+y^2)(1/(x^2)+1/(y^2))=4a^2 (B) (x^2+y^2)^2(1/(x^2)+1/(y^2))=a^2 (C) (x^2+y^2)^2(1/(x^2)+1/(y^2))=4a^2 (D) (x^2+y^2)(1/(x^2)+1/(y^2))=a^2