If `vec r = 3 hati + 2 hatj - 5 hatk , vec a= 2 hati - hatj + hatk, vec b = hati + 3 hatj - 2hatk` `and vec c=-2 hati + hatj - 3 hatk " such that " hat r = x vec a +y vec b + z vec c` then

Options

A. x, y, z are in AP
B. x, y, z are in GP
C. x, y, z are in HP
D. `y, (x)/(2), z ` are in AP

Explanation

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

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

We need to find \(x\), \(y\), and \(z\) such that:

\[ \vec{r} = x\vec{a} + y\vec{b} + z\vec{c} \]

### Step 1: Write the equation in terms of components

Substituting the expressions for \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) into the equation gives:

\[ \vec{r} = x(2\hat{i} - \hat{j} + \hat{k}) + y(\hat{i} + 3\hat{j} - 2\hat{k}) + z(-2\hat{i} + \hat{j} - 3\hat{k}) \]

### Step 2: Combine the components

Now, we combine the components:

\[ \vec{r} = (2x + y - 2z)\hat{i} + (-x + 3y + z)\hat{j} + (x - 2y - 3z)\hat{k} \]

### Step 3: Set up equations by comparing coefficients

Now we can compare the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) from both sides:

1. For \(\hat{i}\):
\[ 2x + y - 2z = 3 \quad \text{(Equation 1)} \]

2. For \(\hat{j}\):
\[ -x + 3y + z = 2 \quad \text{(Equation 2)} \]

3. For \(\hat{k}\):
\[ x - 2y - 3z = -5 \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations

Now we have a system of three equations:

1. \(2x + y - 2z = 3\)
2. \(-x + 3y + z = 2\)
3. \(x - 2y - 3z = -5\)

We can solve these equations simultaneously.

### Step 5: Solve Equation 1 and Equation 2

From Equation 1, we can express \(y\) in terms of \(x\) and \(z\):

\[ y = 3 - 2x + 2z \quad \text{(Equation 4)} \]

Substituting Equation 4 into Equation 2:

\[ -x + 3(3 - 2x + 2z) + z = 2 \]

Expanding this gives:

\[ -x + 9 - 6x + 6z + z = 2 \]
\[ -7x + 7z + 9 = 2 \]
\[ -7x + 7z = -7 \]
\[ x - z = 1 \quad \text{(Equation 5)} \]

### Step 6: Substitute Equation 5 into Equation 3

Now substitute \(z = x - 1\) into Equation 3:

\[ x - 2(3 - 2x + 2(x - 1)) - 3(x - 1) = -5 \]

This simplifies to:

\[ x - 2(3 - 2x + 2x - 2) - 3x + 3 = -5 \]
\[ x - 2(1) - 3x + 3 = -5 \]
\[ -x + 1 = -5 \]
\[ -x = -6 \implies x = 6 \]

### Step 7: Find \(y\) and \(z\)

Using \(x = 6\) in Equation 5:

\[ z = 6 - 1 = 5 \]

Substituting \(x = 6\) into Equation 4 to find \(y\):

\[ y = 3 - 2(6) + 2(5) = 3 - 12 + 10 = 1 \]

### Final Values

Thus, we have:

\[ x = 6, \quad y = 1, \quad z = 5 \]

### Conclusion

The values of \(x\), \(y\), and \(z\) are \(6\), \(1\), and \(5\) respectively.

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