Consider the equation
`E_(1):vecr xx (2hati-hatj+3hatk)=3hati+hatk` and `E_(2): vecr xx (hati+2hatj-3hatk)=2hati-hatj`, then

Options

A. `E_(1)` represents a line
B. `E_(1)` represents two parallel lines
C. `E_(2)` represents a line
D. `E_(2)` represents two parallel planes

Explanation

To solve the given problem, we need to analyze the equations \( E_1 \) and \( E_2 \) to determine whether they represent lines, parallel lines, or parallel planes.

### Step 1: Write down the equations
The equations given are:
- \( E_1: \vec{r} \times (2\hat{i} - \hat{j} + 3\hat{k}) = 3\hat{i} + \hat{k} \)
- \( E_2: \vec{r} \times (\hat{i} + 2\hat{j} - 3\hat{k}) = 2\hat{i} - \hat{j} \)

### Step 2: Express \( \vec{r} \)
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).

### Step 3: Analyze \( E_1 \)
Substituting \( \vec{r} \) into \( E_1 \):
\[ \vec{r} \times (2\hat{i} - \hat{j} + 3\hat{k}) = (x\hat{i} + y\hat{j} + z\hat{k}) \times (2\hat{i} - \hat{j} + 3\hat{k}) \]
Using the determinant method to compute the cross product:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 2 & -1 & 3 \end{vmatrix} \]
Calculating the determinant:
- The coefficient of \( \hat{i} \) is \( (y \cdot 3 - z \cdot (-1)) = 3y + z \)
- The coefficient of \( \hat{j} \) is \( -(x \cdot 3 - z \cdot 2) = -3x + 2z \)
- The coefficient of \( \hat{k} \) is \( (x \cdot (-1) - y \cdot 2) = -x - 2y \)

Thus, we have:
\[ (3y + z)\hat{i} + (-3x + 2z)(-\hat{j}) + (-x - 2y)\hat{k} = 3\hat{i} + \hat{k} \]
This gives us the equations:
1. \( 3y + z = 3 \)
2. \( -3x + 2z = 0 \)
3. \( -x - 2y = 1 \)

### Step 4: Solve the equations from \( E_1 \)
From equation 2, we can express \( z \) in terms of \( x \):
\[ 2z = 3x \implies z = \frac{3}{2}x \]
Substituting \( z \) into equation 1:
\[ 3y + \frac{3}{2}x = 3 \implies 3y = 3 - \frac{3}{2}x \implies y = 1 - \frac{1}{2}x \]
Substituting \( y \) into equation 3:
\[ -x - 2(1 - \frac{1}{2}x) = 1 \implies -x - 2 + x = 1 \implies -2 = 1 \text{ (inconsistent)} \]
Thus, \( E_1 \) represents two parallel lines.

### Step 5: Analyze \( E_2 \)
Substituting \( \vec{r} \) into \( E_2 \):
\[ \vec{r} \times (\hat{i} + 2\hat{j} - 3\hat{k}) = (x\hat{i} + y\hat{j} + z\hat{k}) \times (\hat{i} + 2\hat{j} - 3\hat{k}) \]
Using the determinant method:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & 2 & -3 \end{vmatrix} \]
Calculating the determinant:
- The coefficient of \( \hat{i} \) is \( (y \cdot (-3) - z \cdot 2) = -3y - 2z \)
- The coefficient of \( \hat{j} \) is \( -(x \cdot (-3) - z \cdot 1) = 3x - z \)
- The coefficient of \( \hat{k} \) is \( (x \cdot 2 - y \cdot 1) = 2x - y \)

Thus, we have:
\[ (-3y - 2z)\hat{i} + (3x - z)(-\hat{j}) + (2x - y)\hat{k} = 2\hat{i} - \hat{j} \]
This gives us the equations:
1. \( -3y - 2z = 2 \)
2. \( 3x - z = -1 \)
3. \( 2x - y = 0 \)

### Step 6: Solve the equations from \( E_2 \)
From equation 3:
\[ y = 2x \]
Substituting \( y \) into equation 1:
\[ -3(2x) - 2z = 2 \implies -6x - 2z = 2 \implies 2z = -6x - 2 \implies z = -3x - 1 \]
Substituting \( z \) into equation 2:
\[ 3x - (-3x - 1) = -1 \implies 3x + 3x + 1 = -1 \implies 6x + 1 = -1 \implies 6x = -2 \implies x = -\frac{1}{3} \]
Thus, substituting \( x \) back gives:
\[ y = 2(-\frac{1}{3}) = -\frac{2}{3}, \quad z = -3(-\frac{1}{3}) - 1 = 0 \]
So, \( E_2 \) represents a single line.

### Conclusion
- \( E_1 \) represents two parallel lines.
- \( E_2 \) represents a single line.