If equation of the plane through the parallel lines
`(x+1)/(3)=(y-2)/(2)=(z)/(1)`
and `(x-3)/(3)=(y+4)/(2)=(z-1)/(1)`
is `ax+by-26z+6=0` , then `a+b` = ___________

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) in the equation of the plane given by \( ax + by - 26z + 6 = 0 \) that contains the two parallel lines.

### Step 1: Identify Points on the Lines
The two parallel lines are given in symmetric form. We can extract points from these lines:

1. For the first line \( \frac{x+1}{3} = \frac{y-2}{2} = \frac{z}{1} \):
- When \( t = 0 \), we have the point \( (-1, 2, 0) \).

2. For the second line \( \frac{x-3}{3} = \frac{y+4}{2} = \frac{z-1}{1} \):
- When \( t = 0 \), we have the point \( (3, -4, 1) \).

### Step 2: Substitute the First Point into the Plane Equation
Substituting the first point \( (-1, 2, 0) \) into the plane equation \( ax + by - 26z + 6 = 0 \):

\[ a(-1) + b(2) - 26(0) + 6 = 0 \]
This simplifies to:
\[ -a + 2b + 6 = 0 \quad \text{(1)} \]

### Step 3: Substitute the Second Point into the Plane Equation
Now substitute the second point \( (3, -4, 1) \) into the plane equation:

\[ a(3) + b(-4) - 26(1) + 6 = 0 \]
This simplifies to:
\[ 3a - 4b - 26 + 6 = 0 \]
or:
\[ 3a - 4b - 20 = 0 \quad \text{(2)} \]

### Step 4: Solve the System of Equations
Now we have a system of two equations:
1. \( -a + 2b + 6 = 0 \) (1)
2. \( 3a - 4b - 20 = 0 \) (2)

From equation (1), we can express \( a \) in terms of \( b \):
\[ -a + 2b + 6 = 0 \implies a = 2b + 6 \]

Substituting \( a \) into equation (2):
\[ 3(2b + 6) - 4b - 20 = 0 \]
Expanding this gives:
\[ 6b + 18 - 4b - 20 = 0 \]
Combining like terms:
\[ 2b - 2 = 0 \implies 2b = 2 \implies b = 1 \]

### Step 5: Find \( a \)
Now substitute \( b = 1 \) back into the expression for \( a \):
\[ a = 2(1) + 6 = 2 + 6 = 8 \]

### Step 6: Calculate \( a + b \)
Finally, we calculate \( a + b \):
\[ a + b = 8 + 1 = 9 \]

Thus, the answer is:
\[ \boxed{9} \]