If the equation of circle circumscribing the quadrilateral formed by the lines in order are
`2x+3y=2,3x-2y=3,x+2y=3and2x-y=1` is given by `x^(2)+y^(2)+lamdax+muy+v=0`. Then the value of `|lamda+2mu+v|` is

Answer

To solve the problem, we need to find the equation of the circle that circumscribes the quadrilateral formed by the given lines and then extract the values of λ, μ, and ν to compute |λ + 2μ + ν|.

### Step-by-Step Solution

1. **Identify the equations of the lines:**
The equations of the lines are:
- Line 1: \(2x + 3y = 2\)
- Line 2: \(3x - 2y = 3\)
- Line 3: \(x + 2y = 3\)
- Line 4: \(2x - y = 1\)

2. **Find the intersection points of the lines:**
We need to find the vertices of the quadrilateral formed by these lines by solving pairs of equations.

- **Intersection of Line 1 and Line 2:**
\[ 2x + 3y = 2 \quad (1) \]
\[ 3x - 2y = 3 \quad (2) \]
Solving these equations, we can multiply (1) by 2 and (2) by 3 to eliminate \(y\):
\[ 4x + 6y = 4 \quad (3) \]
\[ 9x - 6y = 9 \quad (4) \]
Adding (3) and (4):
\[ 13x = 13 \implies x = 1 \]
Substituting \(x = 1\) into (1):
\[ 2(1) + 3y = 2 \implies 3y = 0 \implies y = 0 \]
Thus, the intersection point is \(A(1, 0)\).

- **Intersection of Line 2 and Line 3:**
\[ 3x - 2y = 3 \quad (2) \]
\[ x + 2y = 3 \quad (5) \]
Solving these equations, we can multiply (5) by 2:
\[ 2x + 4y = 6 \quad (6) \]
Adding (2) and (6):
\[ 3x - 2y + 2x + 4y = 3 + 6 \implies 5x + 2y = 9 \]
Solving for \(y\):
\[ 2y = 9 - 5x \implies y = \frac{9 - 5x}{2} \]
Substituting back into (2) gives us the coordinates \(B\).

- **Intersection of Line 3 and Line 4:**
\[ x + 2y = 3 \quad (5) \]
\[ 2x - y = 1 \quad (7) \]
Solving these gives us point \(C\).

- **Intersection of Line 1 and Line 4:**
\[ 2x + 3y = 2 \quad (1) \]
\[ 2x - y = 1 \quad (7) \]
Solving these gives us point \(D\).

After solving all pairs, we find the intersection points:
- \(A(1, 0)\)
- \(B(3/2, 3/4)\)
- \(C(1, 1)\)
- \(D\) (calculated similarly).

3. **Equation of the circumscribing circle:**
The equation of the circle is given as:
\[ x^2 + y^2 + \lambda x + \mu y + \nu = 0 \]

4. **Substituting points into the circle equation:**
Substitute each point into the circle equation to form a system of equations.

- For point \(A(1, 0)\):
\[ 1^2 + 0^2 + \lambda(1) + \mu(0) + \nu = 0 \implies \lambda + \nu + 1 = 0 \quad (i) \]

- For point \(B(3/2, 3/4)\):
\[ \left(\frac{3}{2}\right)^2 + \left(\frac{3}{4}\right)^2 + \lambda\left(\frac{3}{2}\right) + \mu\left(\frac{3}{4}\right) + \nu = 0 \quad (ii) \]

- For point \(C(1, 1)\):
\[ 1^2 + 1^2 + \lambda(1) + \mu(1) + \nu = 0 \implies \lambda + \mu + \nu + 2 = 0 \quad (iii) \]

5. **Solving the system of equations:**
From equations (i), (ii), and (iii), we can solve for λ, μ, and ν.

6. **Calculate |λ + 2μ + ν|:**
Substitute the values of λ, μ, and ν into the expression |λ + 2μ + ν|.

### Final Calculation
After solving the equations, we find:
- λ = -17/8
- μ = -1
- ν = 9/8

Now, calculate:
\[ |λ + 2μ + ν| = |-17/8 + 2(-1) + 9/8| = |-17/8 - 16/8 + 9/8| = |-24/8| = 3 \]

### Conclusion
The final answer is:
\[ \boxed{3} \]