Correct Answer - A
Let D be the common difference of the corresponding A.P. and A be its first term. Then,
`(1)/(a) = A + (p -1) D, (1)/(b) = A + (q -1) D and (1)/(c) = A + (r - 1) D`
Now,
`|(bc,ca,ab),(p,q,r),(1,1,1)|`
`= abc |((1)/(a),(1)/(b),(1)/(c)),(p,q,r),(1,1,1)| " " ["Applying" R_(1) rarr R_(1) ((1)/(ac))]`
`= abc |((1)/(a),(1)/(b),(1)/(c)),(p -1,q -1,r -1),(1,1,1)| " " ["Applying" R_(2) rarr R_(2) - R_(3)]`
`= abc |((1)/(a) - (p -1) D,(1)/(b) - (q -1) D,(1)/(c) - (r -1) D),(p -1,q -1,r -1),(1,1,1)| " " ["Applying " R_(1) rarr R_(1) - DR_(2)]`
`= abc |(A,A,A),(p -1,q -1,r -1),(1,1,1)| = abc xx 0 = 0`