If `a,b,c` are real numbers such that `3(a^(2)+b^(2)+c^(2)+1)=2(a+b+c+ab+bc+ca)`, than `a,b,c`are in

Options

A. AP only
B. GP only
C. GP and AP
D. None of these

Explanation

Given, `3(a^(2)+b^(2)+c^(2)+1)=2(a+b+c+ab+bc+ca)`
`implies 2(a^(2)+b^(2)+c^(2)-ab-bc-ca)+(a^(2)+b^(2)+c^(2)-2a -2b -2c+3)`
`implies {(a-b)^(2)+(b-c)^(2)+(c-a)^(2)}+{(a-1)^(2)+(b-1)^(2)+(c-1)}=0`
`implies a-b=b-c=c-a=0 " and "a-1=b-1=c-1=0`
`implies a=b=c=1`
`implies a,b,c` are in GP and AP.
Hence, ia the correct answer.