To find the coordinates of the foot of the perpendicular from the point \( P(1, -2, 1) \) onto the plane containing the two lines given, we will follow these steps:
### Step 1: Identify the direction ratios of the lines
The first line is given by:
\[
\frac{x-1}{6} = \frac{y-1}{7} = \frac{z-3}{8}
\]
The direction ratios of this line are \( (6, 7, 8) \).
The second line is given by:
\[
\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}
\]
The direction ratios of this line are \( (3, 5, 7) \).
### Step 2: Find the normal vector of the plane
To find the normal vector \( \mathbf{n} \) of the plane containing both lines, we take the cross product of their direction ratios:
\[
\mathbf{n} = (6, 7, 8) \times (3, 5, 7)
\]
Calculating the cross product:
\[
\mathbf{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
6 & 7 & 8 \\
3 & 5 & 7
\end{vmatrix}
\]
Expanding this determinant:
\[
\mathbf{n} = \mathbf{i}(7 \cdot 7 - 8 \cdot 5) - \mathbf{j}(6 \cdot 7 - 8 \cdot 3) + \mathbf{k}(6 \cdot 5 - 7 \cdot 3)
\]
\[
= \mathbf{i}(49 - 40) - \mathbf{j}(42 - 24) + \mathbf{k}(30 - 21)
\]
\[
= \mathbf{i}(9) - \mathbf{j}(18) + \mathbf{k}(9)
\]
Thus, the normal vector is \( (9, -18, 9) \). We can simplify this to \( (1, -2, 1) \).
### Step 3: Write the equation of the plane
The plane can be represented by the equation:
\[
1(x - 1) - 2(y + 2) + 1(z - 3) = 0
\]
Expanding this:
\[
x - 1 - 2y - 4 + z - 3 = 0
\]
\[
x - 2y + z - 8 = 0
\]
Thus, the equation of the plane is:
\[
x - 2y + z = 8
\]
### Step 4: Write the equation of the line perpendicular to the plane
The line from point \( P(1, -2, 1) \) to the plane will have the direction ratios of the normal vector \( (1, -2, 1) \). The parametric equations for this line can be written as:
\[
\frac{x - 1}{1} = \frac{y + 2}{-2} = \frac{z - 1}{1} = t
\]
From this, we can express \( x, y, z \) in terms of \( t \):
\[
x = 1 + t, \quad y = -2 - 2t, \quad z = 1 + t
\]
### Step 5: Substitute into the plane equation
Substituting \( x, y, z \) into the plane equation:
\[
(1 + t) - 2(-2 - 2t) + (1 + t) = 8
\]
Simplifying:
\[
1 + t + 4 + 4t + 1 + t = 8
\]
\[
6 + 6t = 8
\]
\[
6t = 2 \quad \Rightarrow \quad t = \frac{1}{3}
\]
### Step 6: Find the coordinates of the foot of the perpendicular
Substituting \( t = \frac{1}{3} \) back into the parametric equations:
\[
x = 1 + \frac{1}{3} = \frac{4}{3}
\]
\[
y = -2 - 2 \cdot \frac{1}{3} = -2 - \frac{2}{3} = -\frac{8}{3}
\]
\[
z = 1 + \frac{1}{3} = \frac{4}{3}
\]
Thus, the coordinates of the foot of the perpendicular are:
\[
Q\left(\frac{4}{3}, -\frac{8}{3}, \frac{4}{3}\right)
\]
### Step 7: Calculate \( a + b + c \)
Let \( a = \frac{4}{3}, b = -\frac{8}{3}, c = \frac{4}{3} \):
\[
a + b + c = \frac{4}{3} - \frac{8}{3} + \frac{4}{3} = \frac{4 - 8 + 4}{3} = \frac{0}{3} = 0
\]
### Final Answer
The value of \( a + b + c \) is \( \boxed{0} \).
