A straight line L cuts the sides AB, AC, AD of a parallelogram ABCD at `B_(1), C_(1), d_(1)` respectively. If `vec(AB_(1))=lambda_(1)vec(AB), vec(AD_(1))=lambda_(2)vec(AD) and vec(AC_(1))=lambda_(3)vec(AC),

Options

A. `lambda_(1), lambda_(3) and lambda_(2) ` are in AP
B. `lambda_(1), lambda_(3) and lambda_(2)` are in GP
C. `lambda_(1),lambda_(3) and lambda_(2)` are in HP
D. `lambda_(1)+lambda_(2)+lambda_(3)=0`

Answer

To solve the problem step by step, we will analyze the given information about the parallelogram ABCD and the line L that intersects its sides.

### Step 1: Define the vectors
Let:
- \( \vec{AB} = \vec{a} \)
- \( \vec{AD} = \vec{b} \)
- \( \vec{AC} = \vec{a} + \vec{b} \)

### Step 2: Express the intersection points in terms of the vectors
Given:
- \( \vec{AB_1} = \lambda_1 \vec{AB} = \lambda_1 \vec{a} \)
- \( \vec{AD_1} = \lambda_2 \vec{AD} = \lambda_2 \vec{b} \)
- \( \vec{AC_1} = \lambda_3 \vec{AC} = \lambda_3 (\vec{a} + \vec{b}) \)

### Step 3: Find the vector \( \vec{B_1D_1} \)
Using the points \( B_1 \) and \( D_1 \):
\[ \vec{B_1D_1} = \vec{AD_1} - \vec{AB_1} = \lambda_2 \vec{b} - \lambda_1 \vec{a} \]

### Step 4: Find the vector \( \vec{C_1B_1} \)
Using the points \( C_1 \) and \( B_1 \):
\[ \vec{C_1B_1} = \vec{AB_1} - \vec{AC_1} = \lambda_1 \vec{a} - \lambda_3 (\vec{a} + \vec{b}) = (\lambda_1 - \lambda_3) \vec{a} - \lambda_3 \vec{b} \]

### Step 5: Establish collinearity condition
Since \( \vec{B_1D_1} \) and \( \vec{C_1B_1} \) are collinear, we can write:
\[ \vec{C_1B_1} = k \vec{B_1D_1} \]
for some scalar \( k \).

### Step 6: Set up the equations
From the collinearity condition, we get:
\[ (\lambda_1 - \lambda_3) \vec{a} - \lambda_3 \vec{b} = k (\lambda_2 \vec{b} - \lambda_1 \vec{a}) \]

### Step 7: Compare coefficients
Comparing coefficients of \( \vec{a} \) and \( \vec{b} \):
1. Coefficient of \( \vec{a} \):
\[ \lambda_1 - \lambda_3 = -k \lambda_1 \]
Rearranging gives:
\[ \lambda_1 + k \lambda_1 = \lambda_3 \implies \lambda_1(1 + k) = \lambda_3 \implies k = \frac{\lambda_3 - \lambda_1}{\lambda_1} \]

2. Coefficient of \( \vec{b} \):
\[ -\lambda_3 = k \lambda_2 \]
Substituting \( k \):
\[ -\lambda_3 = \frac{\lambda_3 - \lambda_1}{\lambda_1} \lambda_2 \]

### Step 8: Solve for \( \lambda_3 \)
Rearranging gives:
\[ -\lambda_3 \lambda_1 = (\lambda_3 - \lambda_1) \lambda_2 \]
Expanding and rearranging leads to:
\[ \lambda_1 \lambda_3 + \lambda_2 \lambda_3 = \lambda_1 \lambda_2 \]
Factoring out \( \lambda_3 \):
\[ \lambda_3 (\lambda_1 + \lambda_2) = \lambda_1 \lambda_2 \implies \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} \]

### Step 9: Find \( \frac{1}{\lambda_3} \)
Taking the reciprocal:
\[ \frac{1}{\lambda_3} = \frac{\lambda_1 + \lambda_2}{\lambda_1 \lambda_2} \]

### Final Answer
Thus, the value of \( \frac{1}{\lambda_3} \) is:
\[ \frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \]