If x, y, z are
Question:

If $x, y, z$ are in arithmetic progression with common difference $\mathrm{d}, x \neq 3 \mathrm{~d}$, 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,

1. 72

2. 12

3. 36

4. 6

Correct Option: 1

Solution:

$\left|\begin{array}{ccc}3 & 4 \sqrt{2} & x \\ 4 & 5 \sqrt{2} & y \\ 5 & k & z\end{array}\right|=0$

$\mathrm{R}_{2} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}$

$\Rightarrow\left|\begin{array}{ccc}3 & 4 \sqrt{2} & x \\ 0 & k-6 \sqrt{2} & 0 \\ 5 & k & z\end{array}\right|=0$

$\Rightarrow(\mathrm{k}-6 \sqrt{2})(3 \mathrm{z}-5 \mathrm{x})=0$

if $3 z-5 x=0 \Rightarrow 3(x+2 d)-5 x=0$

$\Rightarrow x=3 d$ (Not possible)

$\Rightarrow \mathrm{k}=6 \sqrt{2} \quad \Rightarrow \mathrm{k}^{2}=72$