Consider the family of line `(x+y-1)+lamda(2x+3y-5)=0` and `(3x+2y-4)+mu(x+2y-6)=0` Equation of a straight line that belongs to both the families is:

Options

A. `x-2y-8=0`
B. `x-2y+8=0`
C. `2x-y-8=0`
D. None of these

Answer

To find the equation of a straight line that belongs to both families given by the equations:

1. \((x + y - 1) + \lambda(2x + 3y - 5) = 0\)
2. \((3x + 2y - 4) + \mu(x + 2y - 6) = 0\)

we will follow these steps:

### Step 1: Rearranging the First Family of Lines
Start with the first equation:
\[ x + y - 1 + \lambda(2x + 3y - 5) = 0 \]
Rearranging gives:
\[ x + y - 1 + \lambda(2x + 3y - 5) = 0 \]
This can be expressed as:
\[ x + y + \lambda(2x + 3y) - \lambda \cdot 5 - 1 = 0 \]

### Step 2: Finding a Point on the First Line
Set \(\lambda = 0\) to find a specific line:
\[ x + y - 1 = 0 \implies y = -x + 1 \]
To find a point, let’s choose \(x = -2\):
\[ y = -(-2) + 1 = 3 \]
Thus, the point on the first line is \((-2, 3)\).

### Step 3: Rearranging the Second Family of Lines
Now consider the second equation:
\[ (3x + 2y - 4) + \mu(x + 2y - 6) = 0 \]
Rearranging gives:
\[ 3x + 2y - 4 + \mu(x + 2y - 6) = 0 \]

### Step 4: Finding a Point on the Second Line
Set \(\mu = 0\) to find a specific line:
\[ 3x + 2y - 4 = 0 \implies 2y = 4 - 3x \implies y = 2 - \frac{3}{2}x \]
To find a point, let’s choose \(x = -1\):
\[ y = 2 - \frac{3}{2}(-1) = 2 + \frac{3}{2} = \frac{7}{2} \]
Thus, the point on the second line is \((-1, \frac{7}{2})\).

### Step 5: Finding the Equation of the Line through Two Points
Now we have two points:
- Point 1: \((-2, 3)\)
- Point 2: \((-1, \frac{7}{2})\)

Using the two-point form of the equation of a line:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting the points:
\[ y - 3 = \frac{\frac{7}{2} - 3}{-1 - (-2)}(x + 2) \]
Calculating the slope:
\[ \frac{\frac{7}{2} - 3}{-1 + 2} = \frac{\frac{7}{2} - \frac{6}{2}}{1} = \frac{1}{2} \]
Thus, the equation becomes:
\[ y - 3 = \frac{1}{2}(x + 2) \]

### Step 6: Simplifying the Equation
Multiply through by 2 to eliminate the fraction:
\[ 2(y - 3) = x + 2 \]
This simplifies to:
\[ 2y - 6 = x + 2 \implies x - 2y + 8 = 0 \]

### Final Answer
The equation of the straight line that belongs to both families is:
\[ \boxed{x - 2y + 8 = 0} \]