10 different vectors are lying on a plane out of which four are parallel with respect to each other. Probability that three vectors chosen from them will satisfy the equation `lambda_1a+lambda_2b+lambda_3c=0,` where `lambda_1, lambda_2 and lambda_3ne=0` is

Options

A. (a)`(.^6C_2times.^4C_1)/(.^(10)C_3)`
B. (b)`((.^6C_3times.^4C_1)+,^6C_3)/(.^(10)C_3)`
C. (c)`((.^6C_3+times.^4C_1)+,^4C_3)/(.^(10)C_3)`
D. (d)`((.^6C_3 +.^4C_1)+,^6C_2times.^4C_1)/(.^(10)C_3)`

Answer

To solve the problem, we need to find the probability that three vectors chosen from a set of 10 vectors (where 4 are parallel) will satisfy the equation \( \lambda_1 a + \lambda_2 b + \lambda_3 c = 0 \) with \( \lambda_1, \lambda_2, \lambda_3 \neq 0 \).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- We have 10 different vectors in total.
- Out of these, 4 vectors are parallel to each other. Let's denote these parallel vectors as \( A_1, A_2, A_3, A_4 \).
- The remaining 6 vectors are non-parallel and can be denoted as \( B_1, B_2, B_3, B_4, B_5, B_6 \).

2. **Condition for the Equation**:
- The equation \( \lambda_1 a + \lambda_2 b + \lambda_3 c = 0 \) implies that the vectors \( a, b, c \) must be linearly dependent.
- For three vectors to be linearly dependent, at least one of them must be a linear combination of the others. This can happen in the following scenarios:
- All three vectors are from the set of parallel vectors.
- Two vectors are from the parallel set and one from the non-parallel set.
- One vector is from the parallel set and two from the non-parallel set.
- All three vectors are from the non-parallel set.

3. **Calculating the Favorable Outcomes**:
- **Case 1**: All three vectors from the parallel set:
- Choose 3 from 4 parallel vectors: \( \binom{4}{3} = 4 \).

- **Case 2**: Two vectors from the parallel set and one from the non-parallel set:
- Choose 2 from 4 parallel vectors and 1 from 6 non-parallel vectors:
\[ \binom{4}{2} \times \binom{6}{1} = 6 \times 6 = 36. \]

- **Case 3**: One vector from the parallel set and two from the non-parallel set:
- Choose 1 from 4 parallel vectors and 2 from 6 non-parallel vectors:
\[ \binom{4}{1} \times \binom{6}{2} = 4 \times 15 = 60. \]

- **Case 4**: All three vectors from the non-parallel set:
- Choose 3 from 6 non-parallel vectors:
\[ \binom{6}{3} = 20. \]

4. **Total Favorable Outcomes**:
- Adding all the cases together:
\[ 4 + 36 + 60 + 20 = 120. \]

5. **Total Possible Outcomes**:
- The total number of ways to choose 3 vectors from 10:
\[ \binom{10}{3} = 120. \]

6. **Calculating the Probability**:
- The probability \( P \) that three chosen vectors satisfy the condition is given by:
\[ P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{120}{120} = 1. \]

### Final Answer:
The probability that three vectors chosen from them will satisfy the equation \( \lambda_1 a + \lambda_2 b + \lambda_3 c = 0 \) is \( 1 \).

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