The equation of two straight lines are `(x-1)/2=(y+3)/1=(z-2)/(-3)a n d(x-2)/1=(y-1)/(-3)=(z+3)/2dot` Statement 1: the given lines are coplanar. Statement 2: The equations `2x_1-y_1=1,x_1+3y_1=4a n d3x-1+2y_1=5` are consistent.

Options

A. Statement 1 is true, Statement 2 is also true, Statement-2 is the correct explanation of Statement-1.
B. Statement 1 is true, Statement 2 is also true, Statement-2 is not the correct explanation of Statement-1.
C. Statement 1 is true, Statement 2 is false.
D. Statement 1 is false, Statement 2 is true

Answer

To solve the problem step by step, we need to analyze the equations of the two straight lines and determine if they are coplanar. We will also check the consistency of the given equations.

### Step 1: Write the equations of the lines in parametric form

The first line is given by:
\[ \frac{x-1}{2} = \frac{y+3}{1} = \frac{z-2}{-3} \]
Let \( t \) be the parameter. Then we can express \( x, y, z \) in terms of \( t \):
\[ x = 2t + 1, \quad y = t - 3, \quad z = -3t + 2 \]

The second line is given by:
\[ \frac{x-2}{1} = \frac{y-1}{-3} = \frac{z+3}{2} \]
Let \( s \) be the parameter for the second line. Then we can express \( x, y, z \) in terms of \( s \):
\[ x = s + 2, \quad y = -3s + 1, \quad z = 2s - 3 \]

### Step 2: Set the equations equal to find the point of intersection

To find if the lines intersect, we equate the parametric equations:
1. From \( x \):
\[ 2t + 1 = s + 2 \quad \Rightarrow \quad 2t - s = 1 \quad \text{(Equation 1)} \]

2. From \( y \):
\[ t - 3 = -3s + 1 \quad \Rightarrow \quad t + 3s = 4 \quad \text{(Equation 2)} \]

3. From \( z \):
\[ -3t + 2 = 2s - 3 \quad \Rightarrow \quad 3t + 2s = 5 \quad \text{(Equation 3)} \]

### Step 3: Solve the system of equations

We now have a system of three equations:
1. \( 2t - s = 1 \) (Equation 1)
2. \( t + 3s = 4 \) (Equation 2)
3. \( 3t + 2s = 5 \) (Equation 3)

From Equation 1, we can express \( s \) in terms of \( t \):
\[ s = 2t - 1 \]

Substituting \( s \) into Equation 2:
\[ t + 3(2t - 1) = 4 \\ t + 6t - 3 = 4 \\ 7t = 7 \\ t = 1 \]

Now substitute \( t = 1 \) back into the expression for \( s \):
\[ s = 2(1) - 1 = 1 \]

### Step 4: Check if the values satisfy Equation 3

Now we check if \( t = 1 \) and \( s = 1 \) satisfy Equation 3:
\[ 3(1) + 2(1) = 5 \\ 3 + 2 = 5 \quad \text{(True)} \]

### Step 5: Conclusion about coplanarity

Since we found a common point of intersection for both lines, they are coplanar.

### Step 6: Analyze the consistency of the given equations

The equations given are:
1. \( 2x_1 - y_1 = 1 \)
2. \( x_1 + 3y_1 = 4 \)
3. \( 3x - 1 + 2y_1 = 5 \)

We can write the third equation as:
\[ 3x + 2y_1 = 6 \quad \text{(Equation 4)} \]

Now we can check if these equations are consistent. We can solve the first two equations for \( x_1 \) and \( y_1 \):
From Equation 1:
\[ y_1 = 2x_1 - 1 \]

Substituting into Equation 2:
\[ x_1 + 3(2x_1 - 1) = 4 \\ x_1 + 6x_1 - 3 = 4 \\ 7x_1 = 7 \\ x_1 = 1 \]

Substituting \( x_1 = 1 \) back into Equation 1:
\[ y_1 = 2(1) - 1 = 1 \]

Now substituting \( x_1 = 1 \) and \( y_1 = 1 \) into Equation 4:
\[ 3(1) + 2(1) = 6 \\ 3 + 2 = 5 \quad \text{(False)} \]

### Final Conclusion

The lines are coplanar (Statement 1 is true), but the equations are inconsistent (Statement 2 is false).