If x, y, z are three integers in A.P, lying between 1 and 9 and x51, y41 and z31 are three digits numbers, then the value of $\left|\begin{array}{ccc}5 & 4 & 3 \\ x 51 & y 41 & z 31 \\ x & y & z\end{array}\right|$ is

0

The given matrix is:

$\left|\begin{array}{ccc}5 & 4 & 3 \\ x 51 & y 41 & z 31 \\ x & y & z\end{array}\right|$

$\left|\begin{array}{ccc}5 & 4 & 3 \\ 100 x+50+1 & 100 y+40+1 & 100 z+30+1 \\ x & y & z\end{array}\right|$

Applying, R₂ → R₂ –100R₃ –10R₁

$=\left|\begin{array}{ccc}5 & 4 & 3 \\ 1 & 1 & 1 \\ x & y & z\end{array}\right|$

 Applying, \(C_1 \rightarrow C_1 + C_3 - 2C_2\)

$=\left|\begin{array}{ccc}0 & 4 & 3 \\ 0 & 1 & 1 \\ 0 & y & z\end{array}\right| = 0$ (∵ x + z = 2y)

Hence (3) is the correct answer.