The equation of line passing through `(3,-1,2)` and perpendicular to the lines `vec(r)=(hat(i)+hat(j)-hat(k))+lamda(2hat(i)-2hat(j)+hat(k))` and `vec(r)=(2hat(i)+hat(j)-3hat(k))+mu(hat(i)-2hat(j)+2hat(k))` is

Options

A. `(x+3)/(2)=(y+1)/(3)=(z-2)/(2)`
B. `(x-3)/(3)=(y+1)/(2)=(z-2)/(2)`
C. `(x-3)/(2)=(y+1)/(3)=(z-2)/(2)`
D. `(x-2)/(2)=(y+1)/(2)=(z-2)/(3)`

Explanation

To find the equation of the line that passes through the point \( (3, -1, 2) \) and is perpendicular to the given lines, we can follow these steps:

### Step 1: Identify Direction Vectors of Given Lines
The first line is given by the vector equation:
\[ \vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \]
From this, we can identify the direction vector \( \vec{b_1} \) as:
\[ \vec{b_1} = 2\hat{i} - 2\hat{j} + \hat{k} \]

The second line is given by:
\[ \vec{r} = 2\hat{i} + \hat{j} - 3\hat{k} + \mu(\hat{i} - 2\hat{j} + 2\hat{k}) \]
From this, we can identify the direction vector \( \vec{b_2} \) as:
\[ \vec{b_2} = \hat{i} - 2\hat{j} + 2\hat{k} \]

### Step 2: Find the Cross Product of Direction Vectors
To find a direction vector \( \vec{c} \) for the line we want, we need to take the cross product of \( \vec{b_1} \) and \( \vec{b_2} \):
\[ \vec{c} = \vec{b_1} \times \vec{b_2} \]
Calculating the cross product:
\[ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -2 & 2 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Calculating the determinant:
\[ \vec{c} = \hat{i} \begin{vmatrix} -2 & 1 \\ -2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -2 \\ 1 & -2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -2 & 1 \\ -2 & 2 \end{vmatrix} = (-2)(2) - (1)(-2) = -4 + 2 = -2 \)
2. \( \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = (2)(2) - (1)(1) = 4 - 1 = 3 \)
3. \( \begin{vmatrix} 2 & -2 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (-2)(1) = -4 + 2 = -2 \)

Thus,
\[ \vec{c} = -2\hat{i} - 3\hat{j} - 2\hat{k} \]

### Step 4: Write the Equation of the Line
The line passes through the point \( (3, -1, 2) \) and has the direction vector \( \vec{c} = -2\hat{i} - 3\hat{j} - 2\hat{k} \). The vector equation of the line can be written as:
\[ \vec{r} = (3, -1, 2) + \lambda(-2, -3, -2) \]

### Step 5: Convert to Symmetric Form
The symmetric form of the line can be written as:
\[ \frac{x - 3}{-2} = \frac{y + 1}{-3} = \frac{z - 2}{-2} \]

### Final Answer
The equation of the line is:
\[ \frac{x - 3}{-2} = \frac{y + 1}{-3} = \frac{z - 2}{-2} \]