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Step-by-Step Solution
Step 1: Understand the Given Conditions
• We have three real numbers x, y, and z in arithmetic progression (A.P.) with a common difference d.
• Thus, y is the midpoint of x and z, so 2y = x + z.
• x ≠ 3d is an additional given condition.
• The determinant of the matrix
$
\begin{vmatrix}
3 & 4\sqrt{2} & x \\
4 & 5\sqrt{2} & y \\
5 & k & z
\end{vmatrix}
$
is zero. We need to find the value of k².
Step 2: Write Down the Determinant
Consider the determinant:
$
D = \begin{vmatrix}
3 & 4\sqrt{2} & x \\
4 & 5\sqrt{2} & y \\
5 & k & z
\end{vmatrix}.
$
We are given that $D = 0$.
Step 3: Apply a Row Operation
A useful step to simplify the determinant is to replace the first row using the operation
$R_{1} \to R_{1} + R_{3} - 2R_{2}$.
This operation does not change the value of the determinant.
Performing this on each element of the first row:
New $R_{1}$, first element: $3 + 5 - 2 \cdot 4 = 3 + 5 - 8 = 0.$
New $R_{1}$, second element: $4\sqrt{2} + k - 2 \cdot 5\sqrt{2} = 4\sqrt{2} + k - 10\sqrt{2} = k + (4\sqrt{2} - 10\sqrt{2}) = k - 6\sqrt{2}.$
New $R_{1}$, third element: $x + z - 2y.$ Since $x, y, z$ are in A.P., $x + z = 2y$, so $x + z - 2y = 0.$
Hence, the transformed determinant is:
$
D =
\begin{vmatrix}
0 & (k - 6\sqrt{2}) & 0 \\
4 & 5\sqrt{2} & y \\
5 & k & z
\end{vmatrix}.
$
Given $D=0$, we focus on the remaining non-zero elements.
Step 4: Expand the Simplified Determinant
Since the first and third elements of the new $R_{1}$ are zero, the determinant simplifies upon expansion along the first row:
$
D = 0 \cdot
\begin{vmatrix}
5\sqrt{2} & y \\
k & z
\end{vmatrix}
\;-\;
(k - 6\sqrt{2}) \cdot
\begin{vmatrix}
4 & y \\
5 & z
\end{vmatrix}
\;+\;
0 \cdot
\begin{vmatrix}
4 & 5\sqrt{2} \\
5 & k
\end{vmatrix}.
$
Thus,
$
D = -(k - 6\sqrt{2}) \left(4z - 5y\right).
$
The condition $D=0$ implies:
$
(k - 6\sqrt{2})(4z - 5y) = 0.
$
Step 5: Solve for the Possible Values of k
For the product to be zero, at least one factor must be zero:
Case 1: $k - 6\sqrt{2} = 0 \implies k = 6\sqrt{2}.$
Case 2: $4z - 5y = 0 \implies 4z = 5y.$
However, since $x, y, z$ are in A.P. and $x \neq 3d$, we cannot have $4z = 5y$ satisfy the A.P. conditions consistently under the given restriction. Therefore, the valid solution is:
$
k = 6\sqrt{2}.
$
Step 6: Compute k²
$
k^2 = (6\sqrt{2})^2 = 36 \times 2 = 72.
$
Final Answer
Hence, the value of $k^2$ is 72.
