find x, if
`[x" "-5" "-1][{:(1,0,2),(0,2,1),(2,0,3):}][{:(x),(4),(1):}]=0`

Explanation

To solve the equation given in the question, we need to find the value of \( x \) such that the product of the matrices equals zero. Let's go through the solution step by step.

### Step 1: Define the matrices
We have the first matrix:
\[ A = \begin{bmatrix} x & -5 & -1 \end{bmatrix} \]
The second matrix:
\[ B = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \]
And the third matrix:
\[ C = \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} \]

### Step 2: Multiply the first two matrices
We need to multiply matrix \( A \) and matrix \( B \):
\[ A \cdot B = \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \]
Calculating the product:
- The first element:
\[ x \cdot 1 + (-5) \cdot 0 + (-1) \cdot 2 = x - 2 \]
- The second element:
\[ x \cdot 0 + (-5) \cdot 2 + (-1) \cdot 0 = -10 \]
- The third element:
\[ x \cdot 2 + (-5) \cdot 1 + (-1) \cdot 3 = 2x - 5 - 3 = 2x - 8 \]
Thus, we have:
\[ A \cdot B = \begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix} \]

### Step 3: Multiply the result with the third matrix
Now we multiply the result with matrix \( C \):
\[ \begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} \]
Calculating this product:
\[ (x - 2) \cdot x + (-10) \cdot 4 + (2x - 8) \cdot 1 = x^2 - 2x - 40 + 2x - 8 \]
Simplifying:
\[ x^2 - 2x + 2x - 40 - 8 = x^2 - 48 \]

### Step 4: Set the equation to zero
We set the expression equal to zero:
\[ x^2 - 48 = 0 \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x^2 = 48 \]
Taking the square root of both sides:
\[ x = \pm \sqrt{48} = \pm 4\sqrt{3} \]

### Final Answer
Thus, the values of \( x \) are:
\[ x = 4\sqrt{3} \quad \text{or} \quad x = -4\sqrt{3} \]