If `lamda_1, lamda_2 (lambda_1 > lambda_2)` are two values of `lambda` for which the expression `f(x,y) = x^2 + lambdaxy +y^2 - 5 x - 7y + 6` can be resolved as a product of two linear factors, then the value of `3lamda_1 + 2lamda_2` is

Options

A. 5
B. 10
C. 15
D. 20

Answer

To solve the problem, we need to find the values of \( \lambda_1 \) and \( \lambda_2 \) for which the expression

\[ f(x, y) = x^2 + \lambda xy + y^2 - 5x - 7y + 6 \]

can be factored into two linear factors. This occurs when the determinant of the quadratic form is zero.

### Step 1: Identify the coefficients
The given expression can be compared to the general quadratic form:

\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]

Here, we identify:
- \( A = 1 \)
- \( B = \lambda \)
- \( C = 1 \)
- \( D = -5 \)
- \( E = -7 \)
- \( F = 6 \)

### Step 2: Set up the determinant condition
The condition for the expression to be factored into linear factors is that the determinant \( \Delta \) of the coefficients must equal zero:

\[ \Delta = A C - \left( \frac{B}{2} \right)^2 = 0 \]

Substituting the values we have:

\[ \Delta = 1 \cdot 1 - \left( \frac{\lambda}{2} \right)^2 = 0 \]

### Step 3: Solve for \( \lambda \)
Setting the determinant to zero gives us:

\[ 1 - \frac{\lambda^2}{4} = 0 \]

Rearranging this, we get:

\[ \frac{\lambda^2}{4} = 1 \]

Multiplying both sides by 4:

\[ \lambda^2 = 4 \]

Taking the square root of both sides:

\[ \lambda = 2 \quad \text{or} \quad \lambda = -2 \]

Since \( \lambda_1 > \lambda_2 \), we assign:

\[ \lambda_1 = 2, \quad \lambda_2 = -2 \]

### Step 4: Calculate \( 3\lambda_1 + 2\lambda_2 \)
Now, we need to find the value of \( 3\lambda_1 + 2\lambda_2 \):

\[ 3\lambda_1 + 2\lambda_2 = 3(2) + 2(-2) \]

Calculating this gives:

\[ = 6 - 4 = 2 \]

### Final Answer
Thus, the value of \( 3\lambda_1 + 2\lambda_2 \) is:

\[ \boxed{2} \]