If `f_(r)(x), g_(r)(x), h_(r) (x), r=1, 2, 3` are polynomials in x such that ` f_(r)(a) = g_(r)(a) = h_(r) (a), r=1, 2, 3`
`and " "F(x) =|{:(f_(1)(x)" "f_(2)(x)" "f_(3)(x)),(g_(1)(x)" "g_(2)(x)" "g_(3)(x)),(h_(1)(x)" "h_(2)(x)" "h_(3)(x)):}|`
then F'(x) at x = a is ..... .

Answer

To solve the problem, we need to find the derivative of the matrix function \( F(x) \) at \( x = a \). Given that \( f_r(a) = g_r(a) = h_r(a) \) for \( r = 1, 2, 3 \), we can use this information to compute \( F'(x) \).

### Step-by-Step Solution:

1. **Define the Matrix Function**:
The function \( F(x) \) is defined as:
\[ F(x) = \begin{pmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x) \end{pmatrix} \]

2. **Compute the Derivative**:
To find \( F'(x) \), we differentiate each element of the matrix:
\[ F'(x) = \begin{pmatrix} f_1'(x) & f_2'(x) & f_3'(x) \\ g_1'(x) & g_2'(x) & g_3'(x) \\ h_1'(x) & h_2'(x) & h_3'(x) \end{pmatrix} \]

3. **Evaluate at \( x = a \)**:
Now we evaluate \( F'(x) \) at \( x = a \):
\[ F'(a) = \begin{pmatrix} f_1'(a) & f_2'(a) & f_3'(a) \\ g_1'(a) & g_2'(a) & g_3'(a) \\ h_1'(a) & h_2'(a) & h_3'(a) \end{pmatrix} \]

4. **Use the Given Condition**:
Since \( f_r(a) = g_r(a) = h_r(a) \) for \( r = 1, 2, 3 \), this implies that the values of the first row, second row, and third row are equal at \( x = a \). Therefore, we can denote:
\[ f_1(a) = g_1(a) = h_1(a) = k, \quad f_2(a) = g_2(a) = h_2(a) = k, \quad f_3(a) = g_3(a) = h_3(a) = k \]

5. **Apply the Derivative**:
Since the rows of the matrix \( F(a) \) are identical, the derivative \( F'(a) \) will also reflect this. Thus, the determinant of \( F'(a) \) can be computed.

6. **Calculate the Determinant**:
The determinant of a matrix with two identical rows is zero. Therefore:
\[ \text{det}(F'(a)) = 0 \]

7. **Conclusion**:
Hence, we conclude that:
\[ F'(a) = 0 \]

### Final Answer:
Thus, \( F'(x) \) at \( x = a \) is \( 0 \).