The coordinates of the foot of the perpendicular from the point ` (1,-2,1)` on the plane containing the lines, `(x +1)/(6) = ( y-1)/(7) = (z-3)/(8) and (x-1)/(3) = (y-2)/(5) = (z-3)/(7), ` is :

Answer

To find the coordinates of the foot of the perpendicular from the point \( (1, -2, 1) \) on the plane containing the given lines, we will follow these steps:

### Step 1: Identify the direction ratios of the lines
The equations of the lines are given as:
1. \(\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}\)
2. \(\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}\)

From these equations, we can extract the direction ratios:
- For the first line, the direction ratios are \( (6, 7, 8) \).
- For the second line, the direction ratios are \( (3, 5, 7) \).

### Step 2: Find the normal vector of the plane
The normal vector of the plane can be found by taking the cross product of the direction ratios of the two lines.

Let:
\[ \mathbf{a} = (6, 7, 8) \quad \text{and} \quad \mathbf{b} = (3, 5, 7) \]

The cross product \(\mathbf{n} = \mathbf{a} \times \mathbf{b}\) is calculated as follows:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & 7 & 8 \\ 3 & 5 & 7 \end{vmatrix} \]

Calculating the 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:
\[ \mathbf{n} = (9, -18, 9) \]

### Step 3: Simplify the normal vector
We can simplify the normal vector by dividing by 9:
\[ \mathbf{n} = (1, -2, 1) \]

### Step 4: Find the equation of the plane
Using the point \( (1, 2, 3) \) which lies on the plane (as it is a point on the first line), the equation of the plane can be written as:
\[ 1(x - 1) - 2(y - 2) + 1(z - 3) = 0 \]
Expanding this gives:
\[ x - 1 - 2y + 4 + z - 3 = 0 \]
\[ x - 2y + z = 0 \]

### Step 5: Find the foot of the perpendicular
The foot of the perpendicular from the point \( (1, -2, 1) \) to the plane can be found using the formula:
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \frac{-d}{a^2 + b^2 + c^2} \]
where \( (x_1, y_1, z_1) = (1, -2, 1) \) and the plane equation is \( x - 2y + z = 0 \) giving \( d = 0 \).

Here, \( a = 1, b = -2, c = 1 \):
\[ \frac{x - 1}{1} = \frac{y + 2}{-2} = \frac{z - 1}{1} = \frac{0}{1 + 4 + 1} = 0 \]

This implies:
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ y + 2 = 0 \quad \Rightarrow \quad y = -2 \]
\[ z - 1 = 0 \quad \Rightarrow \quad z = 1 \]

### Final Coordinates
Thus, the coordinates of the foot of the perpendicular are:
\[ (1, -2, 1) \]