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 `2r-s=1,r+3s=4a n d3r+2s=5` are consistent.

Options

A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.
B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.
C. Statement 1 is true and Statement 2 is false.
D. Statement 1 is false and Statement 2 is true.

Explanation

To determine whether the given lines are coplanar and whether the equations are consistent, we can follow these steps:

### Step 1: Write the equations of the lines in parametric form
The equations of the two lines are given as:

1. Line 1: \((x-1)/2 = (y+3)/1 = (z-2)/(-3)\)
2. Line 2: \((x-2)/1 = (y-1)/(-3) = (z+3)/2\)

Let’s express these in parametric form.

For Line 1, let \( r \) be the parameter:
- \( x = 2r + 1 \)
- \( y = r - 3 \)
- \( z = -3r + 2 \)

For Line 2, let \( s \) be the parameter:
- \( x = s + 2 \)
- \( y = -3s + 1 \)
- \( z = 2s - 3 \)

### Step 2: Set the equations equal to each other
To find if the lines intersect, we set the parametric equations equal to each other:

1. \( 2r + 1 = s + 2 \) (Equation 1)
2. \( r - 3 = -3s + 1 \) (Equation 2)
3. \( -3r + 2 = 2s - 3 \) (Equation 3)

### Step 3: Rearranging the equations
From Equation 1:
\[ 2r - s = 1 \quad \text{(Equation 1)} \]

From Equation 2:
\[ r + 3s = 4 \quad \text{(Equation 2)} \]

From Equation 3:
\[ 3r + 2s = 5 \quad \text{(Equation 3)} \]

### Step 4: Solve the equations
Now we will solve the system of equations formed by Equations 1, 2, and 3.

1. From Equation 1: \( s = 2r - 1 \)
2. Substitute \( s \) in Equation 2:
\[ r + 3(2r - 1) = 4 \\ r + 6r - 3 = 4 \\ 7r = 7 \\ r = 1 \]

3. Substitute \( r = 1 \) back into \( s = 2r - 1 \):
\[ s = 2(1) - 1 = 1 \]

4. Now substitute \( r = 1 \) and \( s = 1 \) into Equation 3 to check for consistency:
\[ 3(1) + 2(1) = 5 \\ 3 + 2 = 5 \quad \text{(True)} \]

### Conclusion
Since all equations are satisfied, the equations are consistent.

### Step 5: Check coplanarity
The lines are coplanar if the scalar triple product of the direction vectors and the vector between points on the two lines is zero.

- Direction vector of Line 1: \( \vec{d_1} = (2, 1, -3) \)
- Direction vector of Line 2: \( \vec{d_2} = (1, -3, 2) \)
- Vector between points on the lines (using points from parametric equations):
\[ \vec{A} = (1, -3, 2) \quad \text{(from Line 1 when } r = 0\text{)} \]
\[ \vec{B} = (2, 1, -3) \quad \text{(from Line 2 when } s = 0\text{)} \]
\[ \vec{AB} = \vec{B} - \vec{A} = (2 - 1, 1 + 3, -3 - 2) = (1, 4, -5) \]

Now calculate the scalar triple product:
\[ \vec{d_1} \cdot (\vec{d_2} \times \vec{AB}) = 0 \]
If this equals zero, the lines are coplanar.

### Final Answer
- Statement 1: The given lines are coplanar (True).
- Statement 2: The equations are consistent (True).