Consider a plane `prod:vecr*(2hati+hatj-hatk)=5`, a line `L_(1): vecr=(3hati-hatj+2hatk)+lambda(2hati-3hatj-hatk)` and a point `a(3, -4, 1)*L_(2)` is a line passing through A intersecting `L_(1)` and parallel to plane `prod`.
Q. Equation of `L_(2)` is :

Options

A. `vecr=(1+lambda)hati+(2-3lambda)hatj+(1-lambda)hatk:lambda in R`
B. `vecr=(3+lambda)hati-(4-2lambda)hatj+(1+3lambda)hatk, lambda in R`
C. `vecr=(3+lambda)hati-(4+3lambda)hatj+(1-lambda)hatk, lambda in R`
D. None of the above

Answer

To find the equation of the line \( L_2 \) that passes through the point \( A(3, -4, 1) \), intersects the line \( L_1 \), and is parallel to the given plane, we can follow these steps:

### Step 1: Write the equation of line \( L_1 \)

The equation of line \( L_1 \) is given as:
\[ \vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + \lambda(2\hat{i} - 3\hat{j} - \hat{k}) \]
This can be rewritten as:
\[ \vec{r} = (3 + 2\lambda)\hat{i} + (-1 - 3\lambda)\hat{j} + (2 - \lambda)\hat{k} \]

### Step 2: Identify the point on line \( L_2 \)

The line \( L_2 \) passes through the point \( A(3, -4, 1) \). Therefore, we can express \( L_2 \) in vector form as:
\[ \vec{r} = (3\hat{i} - 4\hat{j} + 1\hat{k}) + \mu(a\hat{i} + b\hat{j} + c\hat{k}) \]
where \( a, b, c \) are the direction ratios of line \( L_2 \) and \( \mu \) is a parameter.

### Step 3: Set up the intersection condition

Since \( L_2 \) intersects \( L_1 \), we can equate the coordinates from both lines.

From \( L_1 \):
- \( x = 3 + 2\lambda \)
- \( y = -1 - 3\lambda \)
- \( z = 2 - \lambda \)

From \( L_2 \):
- \( x = 3 + \mu a \)
- \( y = -4 + \mu b \)
- \( z = 1 + \mu c \)

### Step 4: Equate the coordinates

We can set up the following equations based on the coordinates:

1. \( 3 + 2\lambda = 3 + \mu a \) (from \( x \))
2. \( -1 - 3\lambda = -4 + \mu b \) (from \( y \))
3. \( 2 - \lambda = 1 + \mu c \) (from \( z \))

### Step 5: Solve for \( \lambda \) and \( \mu \)

From the first equation:
\[ 2\lambda = \mu a \quad \Rightarrow \quad \mu = \frac{2\lambda}{a} \quad \text{(Equation 1)} \]

From the second equation:
\[ -1 - 3\lambda = -4 + \mu b \quad \Rightarrow \quad \mu b = 3 - 3\lambda \quad \Rightarrow \quad \mu = \frac{3 - 3\lambda}{b} \quad \text{(Equation 2)} \]

From the third equation:
\[ 2 - \lambda = 1 + \mu c \quad \Rightarrow \quad \mu c = 1 - \lambda \quad \Rightarrow \quad \mu = \frac{1 - \lambda}{c} \quad \text{(Equation 3)} \]

### Step 6: Equate the expressions for \( \mu \)

Now we have three expressions for \( \mu \):
1. \( \frac{2\lambda}{a} \)
2. \( \frac{3 - 3\lambda}{b} \)
3. \( \frac{1 - \lambda}{c} \)

Setting Equation 1 equal to Equation 2:
\[ \frac{2\lambda}{a} = \frac{3 - 3\lambda}{b} \]

Cross-multiplying gives:
\[ 2\lambda b = a(3 - 3\lambda) \quad \Rightarrow \quad 2\lambda b + 3a\lambda = 3a \quad \Rightarrow \quad \lambda(2b + 3a) = 3a \quad \Rightarrow \quad \lambda = \frac{3a}{2b + 3a} \]

Setting Equation 1 equal to Equation 3:
\[ \frac{2\lambda}{a} = \frac{1 - \lambda}{c} \]

Cross-multiplying gives:
\[ 2\lambda c = a(1 - \lambda) \quad \Rightarrow \quad 2\lambda c + a\lambda = a \quad \Rightarrow \quad \lambda(2c + a) = a \quad \Rightarrow \quad \lambda = \frac{a}{2c + a} \]

### Step 7: Use the plane condition

The plane is given by:
\[ \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) = 5 \]
For line \( L_2 \) to be parallel to the plane, the direction ratios \( (a, b, c) \) must satisfy:
\[ 2a + b - c = 0 \quad \text{(Equation 4)} \]

### Step 8: Substitute \( b = 3c \)

From the previous steps, we found that \( b = 3c \). Substituting this into Equation 4 gives:
\[ 2a + 3c - c = 0 \quad \Rightarrow \quad 2a + 2c = 0 \quad \Rightarrow \quad a = -c \]

### Step 9: Write the final equation of \( L_2 \)

Substituting \( a = -c \) and \( b = 3c \) into the equation of line \( L_2 \):
\[ \vec{r} = (3\hat{i} - 4\hat{j} + 1\hat{k}) + \mu(-c\hat{i} + 3c\hat{j} + c\hat{k}) \]
Factoring out \( c \):
\[ \vec{r} = (3 - \mu c)\hat{i} + (-4 + 3\mu c)\hat{j} + (1 + \mu c)\hat{k} \]

### Final Answer

Thus, the equation of line \( L_2 \) can be expressed as:
\[ \vec{r} = (3 - \lambda)\hat{i} + (-4 + 3\lambda)\hat{j} + (1 + \lambda)\hat{k} \]