Let \(a,b\) and \(c\) are real numbers which are in arithmetic progression. Let \(\triangle=\left|\begin{array}{lll} 2y+4 & 5y+7 & 8y+a\\ 3y+5 & 6y+8 & 9y+b\\ 4y+6& 7y+9& 10y+c\end{array}\right|\) Then

\(\triangle\) is always zero

The correct answer is Option (3) → \(\triangle\) is always zero

\(\triangle=\left|\begin{array}{lll} 2y+4 & 5y+7 & 8y+a\\ 3y+5 & 6y+8 & 9y+b\\ 4y+6& 7y+9& 10y+c\end{array}\right|\)

$R_1→R_1+R_3-2R_2$

$Δ=\left|\begin{array}{lll} 0& 0 & a+c-2b\\ 3y+5 & 6y+8 & 9y+b\\ 4y+6& 7y+9& 10y+c\end{array}\right|$

and, if a, b and c are in A.P.

$⇒b-a=c-b$

$⇒2b=a+c$

$⇒a+c-2b=0$

$∴Δ=\left|\begin{array}{lll} 0& 0 & 0\\ 3y+5 & 6y+8 & 9y+b\\ 4y+6& 7y+9& 10y+c\end{array}\right|=0$