Question:
If $x, y, z$ are in arithmetic progression with common difference $\mathrm{d}, \mathrm{x} \neq 3 \mathrm{~d}$,
and the determinant of the matrix $\left[\begin{array}{ccc}3 & 4 \sqrt{2} & x \\ 4 & 5 \sqrt{2} & y \\ 5 & k & z\end{array}\right]$ is zero,
then the value of $\mathrm{k}^{2}$ is
Correct Option: 1
Solution:
$\left|\begin{array}{ccc}3 & 4 \sqrt{2} & x \\ 4 & 5 \sqrt{2} & y \\ 5 & k & z\end{array}\right|=0$
$R_{2} \rightarrow R_{1}+R_{3}-2 R_{2}$
$\Rightarrow\left|\begin{array}{ccc}3 & 4 \sqrt{2} & x \\ 0 & k-6 \sqrt{2} & 0 \\ 5 & k & z\end{array}\right|=0$
$\Rightarrow(k-6 \sqrt{2})(3 z-5 x)=0$
if $3 z-5 x=0 \Rightarrow 3(x+2 d)-5 x=0$
$\Rightarrow x=3 \mathrm{~d}$ ( Not possible ) $\Rightarrow \mathrm{k}=6 \sqrt{2} \Rightarrow \mathrm{k}^{2}=72$