Statement 1: The lines `(x-1)/1=y/(-1)=(z+1)/1` and `(x-2)/2=(y+1)/2=z/3` are coplanar and the equation of the plnae containing them is `5x+2y-3z-8=0`
Statement 2: The line `(x-2)/1=(y+1)/2=z/3` is perpendicular to the plane `3x+5y+9z-8=0` and parallel to the plane `x+y-z=0`

Options

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

Answer

To solve the given problem, we will analyze both statements step by step.

### Step 1: Analyze Statement 1
We need to check if the lines given by the equations:
1. \((x-1)/1 = y/(-1) = (z+1)/1\)
2. \((x-2)/2 = (y+1)/2 = z/3\)

are coplanar and if the plane equation \(5x + 2y - 3z - 8 = 0\) contains them.

#### Step 1.1: Find Direction Ratios of the Lines
For the first line:
- Direction Ratios (DR) = (1, -1, 1)

For the second line:
- Direction Ratios (DR) = (2, 2, 3)

#### Step 1.2: Find the Normal Vector of the Plane
The normal vector of the plane given by \(5x + 2y - 3z - 8 = 0\) is:
- Normal Vector (N) = (5, 2, -3)

#### Step 1.3: Check if the Lines are Coplanar
To check if the lines are coplanar, we can use the scalar triple product. The lines are coplanar if the following determinant is zero:

\[ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 2 & 3 \\ 5 & 2 & -3 \end{vmatrix} \]

Calculating the determinant:

\[ = 1 \begin{vmatrix} 2 & 3 \\ 2 & -3 \end{vmatrix} - (-1) \begin{vmatrix} 2 & 3 \\ 5 & -3 \end{vmatrix} + 1 \begin{vmatrix} 2 & 2 \\ 5 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(2 \cdot (-3) - 3 \cdot 2 = -6 - 6 = -12\)
2. \(2 \cdot (-3) - 3 \cdot 5 = -6 - 15 = -21\)
3. \(2 \cdot 2 - 2 \cdot 5 = 4 - 10 = -6\)

Now substituting back into the determinant:

\[ = 1(-12) + 1(-21) + 1(-6) = -12 - 21 - 6 = -39 \]

Since the determinant is not zero, the lines are not coplanar.

### Conclusion for Statement 1
Since the lines are not coplanar, Statement 1 is **False**.

### Step 2: Analyze Statement 2
Now we need to check if the line given by:
\((x-2)/1 = (y+1)/2 = z/3\)

is perpendicular to the plane \(3x + 5y + 9z - 8 = 0\) and parallel to the plane \(x + y - z = 0\).

#### Step 2.1: Find Direction Ratios of the Line
For the line:
- Direction Ratios (DR) = (1, 2, 3)

#### Step 2.2: Find Normal Vectors of the Planes
For the first plane \(3x + 5y + 9z - 8 = 0\):
- Normal Vector (N1) = (3, 5, 9)

For the second plane \(x + y - z = 0\):
- Normal Vector (N2) = (1, 1, -1)

#### Step 2.3: Check Perpendicularity
To check if the line is perpendicular to the first plane, we calculate the dot product of the direction ratios of the line and the normal vector of the plane:

\[ (1, 2, 3) \cdot (3, 5, 9) = 1 \cdot 3 + 2 \cdot 5 + 3 \cdot 9 = 3 + 10 + 27 = 40 \]

Since the dot product is not zero, the line is not perpendicular to the plane.

### Conclusion for Statement 2
Since the line is not perpendicular to the plane, Statement 2 is **False**.

### Final Conclusion
Both statements are false.