Two system of rectangular axes have the same origin. IF a plane cuts them at distances a,b,c and a\',b\',c\'` from the origin then `(A) `1/a^2+1/b^2-1/c^2+1/a^(\'2)+1/b\^(\'2)-1/c^(\'2)=0` (B) `1/a^2-1/b^2-1/c^2+1/a^(\'2)-1/b\^(\'2)-1/c^(\'2)=0` (C) `1/a^2+1/b^2+1/c^2-1/a^(\'2)-1/b\^(\'2)-1/c^(\'2)=0` (D) `1/a^2+1/b^2\+1/c^2+1/a^(\'2)+1/b\^(\'2)+1/c^(\'2)=0`

Two systems of rectangular axes have the same origin. If a plane cuts them at distance a ,b ,cand a^prime ,b^(prime),c ' from the origin, then a. 1/(a^2)+1/(b^2)+1/(c^2)+1/(a^('2))+1/(b^('2))+1/(c^('2))=0 b. 1/(a^2)-1/(b^2)-1/(c^2)+1/(a^('2))-1/(b^('2))-1/(c^('2))=0 c. 1/(a^2)+1/(b^2)+1/(c^2)-1/(a^('2))-1/(b^('2))-1/(c^('2))=0 d. 1/(a^2)+1/(b^2)+1/(c^2)+1/(a^('2))+1/(b^('2))+1/(c^('2))=0

Two systems of rectangular axes have the same origin. If a plane cuts them at distances a ,b ,ca n da^(prime),b^(prime),c^(prime) respectively, prove that 1/(a^2)+1/(b^2)+1/(c^2)=1/(a^('2))+1/(b^('2))+1/(c^('2))

If p is the length of perpendicular from the origin onto the plane whose intercepts on the axes area a,b,c then (A) a+b+c=p (B) a^-2+b^-2+c^-2=p^(-2) (C) a^(-1)+b^(-1)+c^(-1)=p^(-1) (D) a^(-1)+b^(-1)+c^(-1)=1

[[a^(2),b^(2),c^(2)(a+1)^(2),(b+1)^(2),(c+1)^(2)(a-1)^(2),(b-1)^(2),(c-1)^(2)]]=k[[a^(2),b^(2),c^(2)a,b,c1,1,1]]

det[[1,a,b-a,1,c-b,-c,1]]=1+a^(2)+b^(2)+c^(2)

If a^(2)+b^(2)-c^(2)=0, then (1)/(log_(c+a)b)+(1)/(log_(c-a)b)

If a, b, c, d are in GP, then prove that 1/((a^(2)+b^(2))), 1/((b^(2)+c^(2))), 1/((c^(2)+d^(2))) are in GP.