In a four-dimensional space where unit vectors along the axes are `hati,hatj,hatk and hatl, and a_(1),a_(2),a_(3),a_(4) ` are four non-zero vectors such that no vector can be expressed as a linear combination of other `(lamda-1) (a_(1)-a_(2))+mu(a_(2)+a_(3))+gamma(a_(3)+a_(4)-2a_(2))+a_(3)+deltaa_(4)=0`, then

Options

A. (a) `lamda=1`
B. (b) `mu=-(2)/(3)`
C. (c) `gamma=(2)/(3)`
D. (d) `delta=(1)/(3)`

Explanation

To solve the given problem, we need to analyze the equation provided and find the values of the parameters \( \lambda, \mu, \gamma, \) and \( \delta \) under the condition that the vectors \( a_1, a_2, a_3, \) and \( a_4 \) are linearly independent.

### Step-by-Step Solution:

1. **Write the given equation**:
\[ (\lambda - 1)(a_1 - a_2) + \mu(a_2 + a_3) + \gamma(a_3 + a_4 - 2a_2) + a_3 + \delta a_4 = 0 \]

2. **Expand the equation**:
\[ (\lambda - 1)a_1 - (\lambda - 1)a_2 + \mu a_2 + \mu a_3 + \gamma a_3 + \gamma a_4 - 2\gamma a_2 + a_3 + \delta a_4 = 0 \]

3. **Combine like terms**:
- Coefficient of \( a_1 \): \( \lambda - 1 \)
- Coefficient of \( a_2 \): \( -(\lambda - 1) + \mu - 2\gamma \)
- Coefficient of \( a_3 \): \( \mu + \gamma + 1 \)
- Coefficient of \( a_4 \): \( \gamma + \delta \)

Thus, we have:
\[ \begin{align*} \text{Coefficient of } a_1: & \quad \lambda - 1 = 0 \\ \text{Coefficient of } a_2: & \quad -(\lambda - 1) + \mu - 2\gamma = 0 \\ \text{Coefficient of } a_3: & \quad \mu + \gamma + 1 = 0 \\ \text{Coefficient of } a_4: & \quad \gamma + \delta = 0 \\ \end{align*} \]

4. **Solve for \( \lambda \)**:
From the first equation:
\[ \lambda - 1 = 0 \implies \lambda = 1 \]

5. **Substitute \( \lambda \) into the second equation**:
\[ - (1 - 1) + \mu - 2\gamma = 0 \implies \mu - 2\gamma = 0 \implies \mu = 2\gamma \]

6. **Substitute \( \mu \) into the third equation**:
\[ 2\gamma + \gamma + 1 = 0 \implies 3\gamma + 1 = 0 \implies \gamma = -\frac{1}{3} \]

7. **Find \( \mu \)**:
Using \( \mu = 2\gamma \):
\[ \mu = 2 \left(-\frac{1}{3}\right) = -\frac{2}{3} \]

8. **Find \( \delta \)**:
From the fourth equation \( \gamma + \delta = 0 \):
\[ -\frac{1}{3} + \delta = 0 \implies \delta = \frac{1}{3} \]

### Summary of Results:
- \( \lambda = 1 \)
- \( \mu = -\frac{2}{3} \)
- \( \gamma = -\frac{1}{3} \)
- \( \delta = \frac{1}{3} \)