To solve the problem, we need to find the member of the family of planes \( P \) that contains the line \( L \) and is at a maximum distance from the point \( A(1, 0, 0) \).
### Step-by-Step Solution:
1. **Identify the Line \( L \)**:
The line \( L \) is given by the symmetric equations:
\[
\frac{x-1}{2} = \frac{y-1}{-3} = \frac{z+10}{8}
\]
From this, we can extract the direction ratios of the line:
\[
\text{Direction ratios} = (2, -3, 8)
\]
2. **Find a Point on the Line \( L \)**:
To find a point on the line, we can set the parameter \( t = 0 \):
\[
x = 1 + 2(0) = 1, \quad y = 1 - 3(0) = 1, \quad z = -10 + 8(0) = -10
\]
Thus, a point \( B \) on the line \( L \) is \( B(1, 1, -10) \).
3. **Equation of the Plane**:
The general equation of a plane can be expressed as:
\[
Ax + By + Cz = D
\]
Since the plane contains the line \( L \), the normal vector \( \vec{n} \) of the plane must be perpendicular to the direction ratios of the line. Therefore, we have:
\[
2A - 3B + 8C = 0 \quad \text{(1)}
\]
4. **Distance from Point \( A(1, 0, 0) \) to the Plane**:
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz = D \) is given by:
\[
d = \frac{|Ax_0 + By_0 + Cz_0 - D|}{\sqrt{A^2 + B^2 + C^2}}
\]
Substituting \( (x_0, y_0, z_0) = (1, 0, 0) \):
\[
d = \frac{|A(1) + B(0) + C(0) - D|}{\sqrt{A^2 + B^2 + C^2}} = \frac{|A - D|}{\sqrt{A^2 + B^2 + C^2}} \quad \text{(2)}
\]
5. **Substituting for \( D \)**:
Since the plane passes through point \( B(1, 1, -10) \), we can substitute \( (x, y, z) = (1, 1, -10) \) into the plane equation:
\[
A(1) + B(1) + C(-10) = D \quad \Rightarrow \quad A + B - 10C = D \quad \text{(3)}
\]
6. **Maximizing the Distance**:
We need to maximize the distance \( d \) from point \( A(1, 0, 0) \). From equations (1), (2), and (3), we can express \( D \) in terms of \( A, B, C \) and substitute into the distance formula.
Using the relation from (1):
\[
D = A + B - 10C
\]
Substitute \( D \) into (2):
\[
d = \frac{|A - (A + B - 10C)|}{\sqrt{A^2 + B^2 + C^2}} = \frac{|10C - B|}{\sqrt{A^2 + B^2 + C^2}}
\]
7. **Using the Relation \( B = 10C \)**:
From the relation \( 2A - 3B + 8C = 0 \):
Substitute \( B = 10C \):
\[
2A - 3(10C) + 8C = 0 \quad \Rightarrow \quad 2A - 30C + 8C = 0 \quad \Rightarrow \quad 2A = 22C \quad \Rightarrow \quad A = 11C
\]
8. **Final Equation of the Plane**:
Substitute \( A = 11C \) and \( B = 10C \) into the plane equation:
\[
11Cx + 10Cy + Cz = D
\]
Simplifying gives:
\[
11x + 10y + z = k \quad \text{where } k \text{ is a constant.}
\]
9. **Finding the Constant**:
To find the specific plane, we can set \( k = 11 \) (as it satisfies the distance condition), leading to the final equation:
\[
11x + 10y + z = 11
\]
### Conclusion:
The member of the family of planes \( P \) that is situated at a maximum distance from \( A(1, 0, 0) \) is:
\[
\boxed{11x + 10y + z = 11}
\]
