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(A) Given the determinant is zero:$\begin{vmatrix} 3 & 4\sqrt{2} & x \ 4 & 5\sqrt{2} & y \ 5 & k & z \end{vmatrix} = 0$Since $x, y, z$ are in arithm

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$72$

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(A) Given the determinant is zero:$\begin{vmatrix} 3 & 4\sqrt{2} & x \ 4 & 5\sqrt{2} & y \ 5 & k & z \end{vmatrix} = 0$Since $x, y, z$ are in arithmetic progression,we have $y = x + d$ and $z = x + 2d$.Applying the row operation $R_2 ightarrow R_1 + R_3 - 2R_2$:$R_1 + R_3 = (3+5, 4\sqrt{2}+k, x+z) = (8, 4\sqrt{2}+k, 2x+2d)$$2R_2 = (8, 10\sqrt{2}, 2y) = (8, 10\sqrt{2}, 2x+2d)$Subtracting these,the second row becomes $(0, k - 6\sqrt{2}, 0)$.Expanding along the second row:$-(k - 6\sqrt{2}) \begin{vmatrix} 3 & x \ 5 & z \end{vmatrix} = 0$$-(k - 6\sqrt{2})(3z - 5x) = 0$Case $1$: $3z - 5x = 0$$3(x + 2d) - 5x = 0 \Rightarrow 3x + 6d - 5x = 0 \Rightarrow 6d = 2x \Rightarrow x = 3d$.This is not possible as per the given condition $x 
eq 3d$.Case $2$: $k - 6\sqrt{2} = 0 \Rightarrow k = 6\sqrt{2}$.Therefore,$k^2 = (6\sqrt{2})^2 = 36 	imes 2 = 72$.

If $x, y, z$ are in arithmetic progression with common difference $d$,$x \neq 3d$,and the determinant of the matrix $\begin{bmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{bmatrix}$ is zero,then the value of $k^2$ is ..... .

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