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 (a,b,c) then a+b+c=

Explanation

To find the coordinates of the foot of the perpendicular from the point \( (1, -2, 1) \) on the plane containing the lines given by the equations

\[ \frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8} \quad \text{(Line 1)} \]
and
\[ \frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7} \quad \text{(Line 2)}, \]

we can follow these steps:

### Step 1: Identify Direction Vectors and Points on the Lines

From the equations of the lines, we can extract the direction vectors and points:

- For Line 1, the direction vector \( \mathbf{d_1} = \langle 6, 7, 8 \rangle \) and a point on the line is \( P_1 = (-1, 1, 3) \).
- For Line 2, the direction vector \( \mathbf{d_2} = \langle 3, 5, 7 \rangle \) and a point on the line is \( P_2 = (1, 2, 3) \).

### Step 2: Find the Normal Vector to the Plane

The normal vector \( \mathbf{n} \) to the plane containing both lines can be found using the cross product of the direction vectors:

\[ \mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & 7 & 8 \\ 3 & 5 & 7 \end{vmatrix} \]

Calculating this determinant:

\[ \mathbf{n} = \mathbf{i}(7 \cdot 7 - 8 \cdot 5) - \mathbf{j}(6 \cdot 7 - 8 \cdot 3) + \mathbf{k}(6 \cdot 5 - 7 \cdot 3) \]

Calculating each component:

- \( \mathbf{i} \): \( 49 - 40 = 9 \)
- \( \mathbf{j} \): \( 42 - 24 = 18 \) (note the negative sign in front)
- \( \mathbf{k} \): \( 30 - 21 = 9 \)

Thus,

\[ \mathbf{n} = \langle 9, -18, 9 \rangle \]

### Step 3: Equation of the Plane

Using the normal vector and a point on the plane (we can use either \( P_1 \) or \( P_2 \)), we can write the equation of the plane. Using point \( P_1 = (-1, 1, 3) \):

\[ 9(x + 1) - 18(y - 1) + 9(z - 3) = 0 \]

Expanding this gives:

\[ 9x + 9 - 18y + 18 + 9z - 27 = 0 \implies 9x - 18y + 9z = 0 \]

### Step 4: 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 for the foot of the perpendicular from a point to a plane. The general formula is:

\[ \text{Foot} = \left( x_0, y_0, z_0 \right) - \frac{d}{|\mathbf{n}|} \mathbf{n} \]

where \( d \) is the distance from the point to the plane, and \( |\mathbf{n}| \) is the magnitude of the normal vector.

Calculating \( d \):

Substituting \( (1, -2, 1) \) into the plane equation:

\[ 9(1) - 18(-2) + 9(1) = 9 + 36 + 9 = 54 \]

Thus,

\[ d = \frac{54}{\sqrt{9^2 + (-18)^2 + 9^2}} = \frac{54}{\sqrt{81 + 324 + 81}} = \frac{54}{\sqrt{486}} = \frac{54}{9\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6} \]

Now, substituting back to find the foot of the perpendicular:

\[ \text{Foot} = (1, -2, 1) - \frac{\sqrt{6}}{9} \langle 9, -18, 9 \rangle \]

Calculating this gives:

\[ \text{Foot} = \left( 1 - \sqrt{6}, -2 + 2\sqrt{6}, 1 - \sqrt{6} \right) \]

### Step 5: Calculate \( a + b + c \)

Let \( a = 1 - \sqrt{6} \), \( b = -2 + 2\sqrt{6} \), \( c = 1 - \sqrt{6} \).

Thus,

\[ a + b + c = (1 - \sqrt{6}) + (-2 + 2\sqrt{6}) + (1 - \sqrt{6}) = 0 \]

### Final Answer

The value of \( a + b + c \) is:

\[ \boxed{0} \]