(A) `4a^(2)+9b^(2)+16c^(2)=2(3ab+6bc+4ca)`
`(2a)^(2)+(3b)^(2)+(4c)^(2)-(2a)(3b)(4c)-(2a)(4c)=0`
`(1)/(2){(2a-3b)^(2)+(3b-4c)^(2)+(4c-2a)^(2)}=0`
`implies 2a-3b=0 " and " 3b-4c=0`
and `4c-2a=0 implies 2a=3b`
and `3b=4c " and " 4c=2a`
`implies a=(3)/(2)b " and "b=(4)/(3)c " and " c=(1)/(2)a`
`implies a=(3)/(2)b " and "b=(4)/(3)c " and " c=(3)/(4)b`
So, a,b,c are `(3)/(2)b,b,(3)/(4)b`
Reciprocal of the terms `(2)/(3b),(1)/(b),(4)/(3b)`.
which is in AP.
So, these a,b,c are in HP.
(B) `17a^(2)+13b^(2)+5c^(2)=3ab+15bc+5ca`
`implies 34a^(2)+26b^(2)+10c^(2)-6ab-30bc-10ca=0`
`implies (3a-b)^(2)+(5b-3c)^(2)+(c-5a)^(2)=0`
`implies 3a-b=0 " and " 5b-3c=0 " and " c-5a=0`
`implies (a)/(1)=(b)/(3)=(c)/(5)=lambda" " [" say "]`
`:. a=lambda,b=3lambda,c=5lambda`
Hence, a,b,c are in AP.
(C ) `a^(2)+9b^(2)+25c^(2)=abc((15)/(a)+(5)/(b)+(3)/(c))`
`(a)^(2)+(3b)^(2)+(5c)^(2)-15bc-5ac-3ab=0`
`(1)/(2){(a-3b)^(2)+(3b-5c)^(2)+(5c-a)^(2)}=0`
`implies a-3b=0 " and "3b-5c=0 " and " 5c-a=0`
`implies a=3b " and " 3b=5c " and " 5c=a`
`implies a=3b " and " b=(5)/(3)c " and " c=(a)/(5)`
`implies a=3b " and " b=(5)/(3)c " and " c=(3)/(5)b`
So, a,b,c are of the form `3b,b,(3b)/(5)`.
Reciprocal of `3b,b,(3b)/(5) " are " (1)/(3b),(1)/(b),(5)/(3b)`, which are in AP.
`[:.(1)/(b)-(1)/(3b)=(2)/(3b) " annd " (5)/(3b)-(1)/(b)=(2)/(3b)]`
(D) `(a^(2)+b^(2)+c^(2))p^(2)-2p(qb+bc+ca)+a^(2)+b^(2)+c^(2)le0`
`implies (a^(2)p^(2)+b^(2)-2abp)+(b^(2)p^(2)+c^(2)-2pbc)+(c^(2)p^(2)+a^(2)-2acp)le0`
`implies (ap-b)^(2)+(bp-c)^(2)+(cp-a)^(2)le0`
`implies (ap-b)^(2)+(bp-c)^(2)+(cp-a)^(2)=0`
`implies ap-b=0 " and " bp-c=0 " and " cp-a=0`
`implies p=(b)/(a) " and " p=(c)/(b) " and " p=(a)/(c)`
`implies (b)/(a)=(c)/(b)=(a)/(c)`
implies a,b,c are in GP.
