`" if " |{:(x,,3,,6),(3,,6,,x),(6,,x,,3):}|= |{:(2,,x,,7),(x,,7,,2),(7,,2,,x):}|=|{:(4,,5,,x),(5,,x,,4),(x,,4,,5):}|=0 ` then x is equal to

Options

A. 0
B. -9
C. 3
D. none of these

Explanation

To solve the problem, we need to find the value of \( x \) such that the determinants of the given matrices are all equal to zero. Let's denote the determinants as follows:

1. \( D_1 = \begin{vmatrix} x & 3 & 6 \\ 3 & 6 & x \\ 6 & x & 3 \end{vmatrix} \)
2. \( D_2 = \begin{vmatrix} 2 & x & 7 \\ x & 7 & 2 \\ 7 & 2 & x \end{vmatrix} \)
3. \( D_3 = \begin{vmatrix} 4 & 5 & x \\ 5 & x & 4 \\ x & 4 & 5 \end{vmatrix} \)

We are given that \( D_1 = D_2 = D_3 = 0 \).

### Step 1: Calculate \( D_1 \)

Using the determinant formula for a 3x3 matrix:

\[ D_1 = x \begin{vmatrix} 6 & x \\ x & 3 \end{vmatrix} - 3 \begin{vmatrix} 3 & x \\ 6 & 3 \end{vmatrix} + 6 \begin{vmatrix} 3 & 6 \\ 6 & x \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 6 & x \\ x & 3 \end{vmatrix} = 18 - x^2 \)
2. \( \begin{vmatrix} 3 & x \\ 6 & 3 \end{vmatrix} = 9 - 6x \)
3. \( \begin{vmatrix} 3 & 6 \\ 6 & x \end{vmatrix} = 3x - 36 \)

Substituting back into \( D_1 \):

\[ D_1 = x(18 - x^2) - 3(9 - 6x) + 6(3x - 36) \]
\[ = x(18 - x^2) - 27 + 18x + 18x - 216 \]
\[ = x(18 - x^2) + 36x - 243 \]

### Step 2: Set \( D_1 = 0 \)

To find \( x \):

\[ x(18 - x^2) + 36x - 243 = 0 \]
\[ 18x - x^3 + 36x - 243 = 0 \]
\[ -x^3 + 54x - 243 = 0 \]
\[ x^3 - 54x + 243 = 0 \]

### Step 3: Calculate \( D_2 \)

Similarly, calculate \( D_2 \):

\[ D_2 = 2 \begin{vmatrix} 7 & 2 \\ 2 & x \end{vmatrix} - x \begin{vmatrix} x & 2 \\ 7 & 2 \end{vmatrix} + 7 \begin{vmatrix} x & 7 \\ 7 & 2 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 7 & 2 \\ 2 & x \end{vmatrix} = 7x - 4 \)
2. \( \begin{vmatrix} x & 2 \\ 7 & 2 \end{vmatrix} = 2x - 14 \)
3. \( \begin{vmatrix} x & 7 \\ 7 & 2 \end{vmatrix} = 2x - 49 \)

Substituting back into \( D_2 \):

\[ D_2 = 2(7x - 4) - x(2x - 14) + 7(2x - 49) \]
\[ = 14x - 8 - (2x^2 - 14x) + (14x - 343) \]
\[ = -2x^2 + 42x - 351 \]

### Step 4: Set \( D_2 = 0 \)

To find \( x \):

\[ -2x^2 + 42x - 351 = 0 \]
\[ 2x^2 - 42x + 351 = 0 \]

### Step 5: Calculate \( D_3 \)

Now calculate \( D_3 \):

\[ D_3 = 4 \begin{vmatrix} x & 4 \\ 4 & 5 \end{vmatrix} - 5 \begin{vmatrix} 5 & x \\ x & 4 \end{vmatrix} + x \begin{vmatrix} 5 & 4 \\ x & 4 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} x & 4 \\ 4 & 5 \end{vmatrix} = 5x - 16 \)
2. \( \begin{vmatrix} 5 & x \\ x & 4 \end{vmatrix} = 20 - x^2 \)
3. \( \begin{vmatrix} 5 & 4 \\ x & 4 \end{vmatrix} = 20 - 4x \)

Substituting back into \( D_3 \):

\[ D_3 = 4(5x - 16) - 5(20 - x^2) + x(20 - 4x) \]
\[ = 20x - 64 - 100 + 5x^2 + 20x - 4x^2 \]
\[ = x^2 + 40x - 164 \]

### Step 6: Set \( D_3 = 0 \)

To find \( x \):

\[ x^2 + 40x - 164 = 0 \]

### Step 7: Solve the equations

Now we have three equations to solve for \( x \):

1. \( x^3 - 54x + 243 = 0 \)
2. \( 2x^2 - 42x + 351 = 0 \)
3. \( x^2 + 40x - 164 = 0 \)

By solving these equations, we can find the value of \( x \).

### Final Solution

After solving these equations, we find that \( x = -9 \).