If x, y, z are in A.P., then the value of the det A. ;Where ;A=$\begin{vmatrix} 4 &5 & 6 & x\\ 5& 6 & 7 &y \\ 6& 7& 8 &z \\ x & y & z & 0 \end{vmatrix}$ 

$|\mathrm{A}|=\left|\begin{array}{llll}4 & 5 & 6 & \mathrm{x} \\ 5 & 6 & 7 & \mathrm{y} \\ 6 & 7 & 8 & \mathrm{z} \\ \mathrm{x} & \mathrm{y} & \mathrm{z} & 0\end{array}\right|$

$=-x\left|\begin{array}{lll}5 & 6 & x \\ 6 & 7 & y \\ 7 & 8 & z\end{array}\right|+y\left|\begin{array}{lll}4 & 6 & x \\ 5 & 7 & y \\ 6 & 8 & z\end{array}\right|-z\left|\begin{array}{lll}4 & 5 & x \\ 5 & 6 & y \\ 6 & 7 & z\end{array}\right|+0\left|\begin{array}{lll}4 & 5 & 6 \\ 5 & 6 & 7 \\ 6 & 7 & 8\end{array}\right|$

$\begin{array}{l} =-x(0)+y(0)-z(0)+0 \\ \end{array}$

Determinate value of a matrix is zero if all of its rows or column are in A.P. In all above $3 \times 3$ determinate, each column is A.P.

$\Rightarrow|\mathrm{A}|=0$