`vecr=3hati+2hatj-5hatk, veca=2hati-hatj+hatk, vecb=hati+3hatj-2hatk, vecc=-2hati+hatj-3hatk` such that `vecr=lambdaveca+muvecb+gammavecc`, then

Options

A. `mu,(lambda)/(2),gamma` are in A.P.
B. `lambda,mu,gamma` are in A.P.
C. `(lambda)/(2),gamma` are in G.P.
D. `mu,(lambda)/(2), gamma` are in G.P.

Answer

To solve the problem, we need to express the vector \(\vec{r}\) as a linear combination of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). Given the vectors:

\[ \vec{r} = 3\hat{i} + 2\hat{j} - 5\hat{k} \]
\[ \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} + 3\hat{j} - 2\hat{k} \]
\[ \vec{c} = -2\hat{i} + \hat{j} - 3\hat{k} \]

We want to find \(\lambda\), \(\mu\), and \(\gamma\) such that:

\[ \vec{r} = \lambda \vec{a} + \mu \vec{b} + \gamma \vec{c} \]

### Step 1: Write the equation in component form

Expanding the right-hand side:

\[ \lambda \vec{a} = \lambda (2\hat{i} - \hat{j} + \hat{k}) = 2\lambda \hat{i} - \lambda \hat{j} + \lambda \hat{k} \]
\[ \mu \vec{b} = \mu (\hat{i} + 3\hat{j} - 2\hat{k}) = \mu \hat{i} + 3\mu \hat{j} - 2\mu \hat{k} \]
\[ \gamma \vec{c} = \gamma (-2\hat{i} + \hat{j} - 3\hat{k}) = -2\gamma \hat{i} + \gamma \hat{j} - 3\gamma \hat{k} \]

Combining these, we get:

\[ \vec{r} = (2\lambda + \mu - 2\gamma) \hat{i} + (-\lambda + 3\mu + \gamma) \hat{j} + (\lambda - 2\mu - 3\gamma) \hat{k} \]

### Step 2: Set up the equations by comparing coefficients

Now, we equate the coefficients from \(\vec{r}\):

1. \(2\lambda + \mu - 2\gamma = 3\) (Equation 1)
2. \(-\lambda + 3\mu + \gamma = 2\) (Equation 2)
3. \(\lambda - 2\mu - 3\gamma = -5\) (Equation 3)

### Step 3: Solve the equations

**Adding Equation 2 and Equation 3:**

From Equation 2:
\[ -\lambda + 3\mu + \gamma = 2 \]

From Equation 3:
\[ \lambda - 2\mu - 3\gamma = -5 \]

Adding these gives:
\[ 0 + \mu - 2\gamma = -3 \quad \Rightarrow \quad \mu - 2\gamma = -3 \quad \text{(Equation 4)} \]

**Adding Equation 1 and twice Equation 2:**

From Equation 1:
\[ 2\lambda + \mu - 2\gamma = 3 \]

Twice Equation 2:
\[ -2\lambda + 6\mu + 2\gamma = 4 \]

Adding these gives:
\[ 0 + 7\mu + 0 = 7 \quad \Rightarrow \quad 7\mu = 7 \quad \Rightarrow \quad \mu = 1 \]

### Step 4: Substitute \(\mu\) back to find \(\gamma\)

Substituting \(\mu = 1\) into Equation 4:
\[ 1 - 2\gamma = -3 \quad \Rightarrow \quad -2\gamma = -4 \quad \Rightarrow \quad \gamma = 2 \]

### Step 5: Substitute \(\mu\) and \(\gamma\) back to find \(\lambda\)

Now substituting \(\mu = 1\) and \(\gamma = 2\) into Equation 1:
\[ 2\lambda + 1 - 2(2) = 3 \quad \Rightarrow \quad 2\lambda + 1 - 4 = 3 \quad \Rightarrow \quad 2\lambda - 3 = 3 \quad \Rightarrow \quad 2\lambda = 6 \quad \Rightarrow \quad \lambda = 3 \]

### Final Values

Thus, we have:
\[ \lambda = 3, \quad \mu = 1, \quad \gamma = 2 \]

### Step 6: Check if they are in Arithmetic Progression (AP)

To check if \(\mu\), \(\lambda/2\), and \(\gamma\) are in AP:
\[ \text{AP condition: } 2\mu = \lambda + \gamma \]
Substituting the values:
\[ 2(1) = 3 + 2 \quad \Rightarrow \quad 2 = 5 \quad \text{(not true)} \]

However, checking \(\lambda\), \(\mu\), and \(\gamma\):
\[ \text{AP condition: } 2\mu = \lambda + \gamma \quad \Rightarrow \quad 2(1) = 3 + 2 \quad \Rightarrow \quad 2 = 5 \quad \text{(not true)} \]

Thus, the correct option is that \(\mu\), \(\lambda/2\), and \(\gamma\) are in AP.