Consider three planes `P_(1):x-y+z=1`, `P_(2):x+y-z=-1` and
`P_(3):x-3y+3z=2`. Let `L_(1),L_(2),L_(3)` be the lines of intersection of the planes `P_(2)` and `P_(3),P_(3)` and `P_(1),P_(1)` and `P_(2)` respectively.
Statement I Atleast two of the lines `L_(1),L_(2)` and `L_(3)` are non-parallel.
Statement II The three planes do not have a common point.

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 problem, we need to analyze the three planes and their intersections step by step.

### Step 1: Identify the given planes
We have three planes defined by the following equations:
1. \( P_1: x - y + z = 1 \)
2. \( P_2: x + y - z = -1 \)
3. \( P_3: x - 3y + 3z = 2 \)

### Step 2: Find the direction ratios of the lines of intersection
The lines of intersection \( L_1, L_2, L_3 \) are formed by the intersections of the planes:

- \( L_1 \): Intersection of \( P_2 \) and \( P_3 \)
- \( L_2 \): Intersection of \( P_3 \) and \( P_1 \)
- \( L_3 \): Intersection of \( P_1 \) and \( P_2 \)

We will use the formula for the direction ratios of the line formed by the intersection of two planes given by:
\[ \text{Direction Ratios} = (b_1c_2 - b_2c_1, c_1a_2 - c_2a_1, a_1b_2 - a_2b_1) \]
where \( a, b, c \) are the coefficients of \( x, y, z \) in the plane equations.

### Step 3: Calculate direction ratios for \( L_1 \) (intersection of \( P_2 \) and \( P_3 \))
For \( P_2: x + y - z + 1 = 0 \) (coefficients: \( 1, 1, -1 \)) and \( P_3: x - 3y + 3z - 2 = 0 \) (coefficients: \( 1, -3, 3 \)):
- \( b_1 = 1, c_1 = -1 \)
- \( b_2 = -3, c_2 = 3 \)

Calculating:
\[ \text{Direction Ratios of } L_1 = (1 \cdot 3 - 1 \cdot -1, -1 \cdot 1 - 3 \cdot 1, 1 \cdot -3 - 1 \cdot 1) = (3 + 1, -1 - 3, -3 - 1) = (4, -4, -4) \]

### Step 4: Calculate direction ratios for \( L_2 \) (intersection of \( P_3 \) and \( P_1 \))
For \( P_3 \) and \( P_1 \):
- \( P_1: x - y + z - 1 = 0 \) (coefficients: \( 1, -1, 1 \))
- \( P_3: x - 3y + 3z - 2 = 0 \)

Calculating:
\[ \text{Direction Ratios of } L_2 = (-1 \cdot 3 - 1 \cdot 1, 1 \cdot 1 - 3 \cdot 1, 1 \cdot -3 - 1 \cdot -1) = (-3 - 1, 1 - 3, -3 + 1) = (-4, -2, -2) \]

### Step 5: Calculate direction ratios for \( L_3 \) (intersection of \( P_1 \) and \( P_2 \))
For \( P_1 \) and \( P_2 \):
- \( P_1 \) and \( P_2 \)

Calculating:
\[ \text{Direction Ratios of } L_3 = (-1 \cdot -1 - 1 \cdot 1, 1 \cdot 1 - -1 \cdot 1, 1 \cdot 1 - 1 \cdot -1) = (1 - 1, 1 + 1, 1 + 1) = (0, 2, 2) \]

### Step 6: Analyze the direction ratios
Now we have:
- \( L_1: (4, -4, -4) \)
- \( L_2: (-4, -2, -2) \)
- \( L_3: (0, 2, 2) \)

To check if at least two lines are non-parallel, we observe that:
- \( L_1 \) and \( L_2 \) are not scalar multiples of each other.
- \( L_3 \) is also not parallel to either \( L_1 \) or \( L_2 \).

### Step 7: Check if the planes have a common point
To check if the three planes intersect at a common point, we can add the equations of the first two planes:
1. \( P_1: x - y + z = 1 \)
2. \( P_2: x + y - z = -1 \)

Adding these gives:
\[ 2x = 0 \implies x = 0 \]

Substituting \( x = 0 \) into \( P_1 \):
\[ 0 - y + z = 1 \implies z = 1 + y \]

Substituting \( x = 0 \) into \( P_2 \):
\[ 0 + y - z = -1 \implies y - (1 + y) = -1 \implies -1 = -1 \]

Now substituting \( x = 0 \) and \( z = 1 + y \) into \( P_3 \):
\[ 0 - 3y + 3(1 + y) = 2 \implies 0 - 3y + 3 + 3y = 2 \implies 3 = 2 \]
This is a contradiction, indicating that there is no common point.

### Conclusion
- Statement I: At least two of the lines \( L_1, L_2, L_3 \) are non-parallel. **True**
- Statement II: The three planes do not have a common point. **True**

### Final Answer
Both statements are true.