Let `vec a=hat i+2hat j+3hat k,vec b=hat i-hat j+2hat k` and `vec c=5hat i-3hat j+3hat k`,be there(three) vector.If `vec r` is a vector such that, `vec r times vec b=vec c times vec b` and `vec r .(vec a)=0,` then `25 abs(vec r)^(2)` is equal to

Options

A. `560`
B. `339`
C. `449`
D. `336`

Answer

To solve the problem, we need to find the vector \(\vec{r}\) that satisfies the conditions given in the question. Let's break down the solution step by step.

### Step 1: Understand the conditions
We have:
1. \(\vec{r} \times \vec{b} = \vec{c} \times \vec{b}\)
2. \(\vec{r} \cdot \vec{a} = 0\)

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

### Step 2: Calculate \(\vec{c} \times \vec{b}\)
First, we need to compute \(\vec{c} \times \vec{b}\):
\[ \vec{c} = \begin{pmatrix} 5 \\ -3 \\ 3 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \]
Using the determinant to find the cross product:
\[ \vec{c} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & -3 & 3 \\ 1 & -1 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}((-3)(2) - (3)(-1)) - \hat{j}((5)(2) - (3)(1)) + \hat{k}((5)(-1) - (-3)(1)) \]
\[ = \hat{i}(-6 + 3) - \hat{j}(10 - 3) + \hat{k}(-5 + 3) \]
\[ = \hat{i}(-3) - \hat{j}(7) + \hat{k}(-2) \]
\[ = -3\hat{i} - 7\hat{j} - 2\hat{k} \]

### Step 3: Set up the equation for \(\vec{r}\)
From the first condition, we have:
\[ \vec{r} \times \vec{b} = -3\hat{i} - 7\hat{j} - 2\hat{k} \]

### Step 4: Express \(\vec{r}\) in terms of a scalar multiple
Let \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\). Then:
\[ \vec{r} \times \vec{b} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \]
Calculating this cross product:
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & -1 & 2 \end{vmatrix} \]
\[ = \hat{i}(y \cdot 2 - z \cdot (-1)) - \hat{j}(x \cdot 2 - z \cdot 1) + \hat{k}(x \cdot (-1) - y \cdot 1) \]
\[ = (2y + z)\hat{i} - (2x - z)\hat{j} + (-x - y)\hat{k} \]

Setting this equal to \(-3\hat{i} - 7\hat{j} - 2\hat{k}\), we get the following equations:
1. \(2y + z = -3\)
2. \(2x - z = 7\)
3. \(-x - y = -2\)

### Step 5: Solve the system of equations
From equation 3:
\[ x + y = 2 \quad \Rightarrow \quad y = 2 - x \]
Substituting \(y\) in equations 1 and 2:
1. \(2(2 - x) + z = -3\) \(\Rightarrow 4 - 2x + z = -3\) \(\Rightarrow z = -7 + 2x\)
2. \(2x - z = 7\) \(\Rightarrow 2x - (-7 + 2x) = 7\) \(\Rightarrow 2x + 7 - 2x = 7\) \(\Rightarrow 7 = 7\) (always true)

Now substituting \(z\) into \(2y + z = -3\):
\[ 2(2 - x) + (-7 + 2x) = -3 \]
\[ 4 - 2x - 7 + 2x = -3 \quad \Rightarrow \quad -3 = -3 \quad \text{(always true)} \]

### Step 6: Choose a value for \(x\)
Let’s choose \(x = 0\):
\[ y = 2 - 0 = 2, \quad z = -7 + 2(0) = -7 \]
Thus, \(\vec{r} = 0\hat{i} + 2\hat{j} - 7\hat{k}\).

### Step 7: Calculate \(|\vec{r}|^2\)
\[ |\vec{r}|^2 = 0^2 + 2^2 + (-7)^2 = 0 + 4 + 49 = 53 \]

### Step 8: Calculate \(25 |\vec{r}|^2\)
\[ 25 |\vec{r}|^2 = 25 \times 53 = 1325 \]

### Final Answer
Thus, \(25 |\vec{r}|^2 = 1325\).