The user wants me to analyze two statements regarding three planes and their lines of intersection.

The given planes are: ${P_1}: x - y + z = 1$ ${P_2}: x + y - z = 1$ ${P_3}: x - 3y + 3z = 2$

Let the normal vectors to the planes be $\vec{n_1}, \vec{n_2}, \vec{n_3}$ respectively. $\vec{n_1} = \langle 1, -1, 1 \rangle$ $\vec{n_2} = \langle 1, 1, -1 \rangle$ $\vec{n_3} = \langle 1, -3, 3 \rangle$

Let ${L_1}, {L_2}, {L_3}$ be the lines of intersection of the planes ${P_2}$ and ${P_3}$, ${P_3}$ and ${P_1}$, and ${P_1}$ and ${P_2}$ respectively.

Analysis of Statement 2: The three planes do not have a common point.

To check for a common point, we need to solve the system of linear equations:

1. $x - y + z = 1$
2. $x + y - z = 1$
3. $x - 3y + 3z = 2$

Step 1: Solve the system of equations. We can add equation (1) and equation (2): $(x - y + z) + (x + y - z) = 1 + 1$ $2x = 2 \implies x = 1$

Substitute $x = 1$ into equations (1) and (2): From (1): $1 - y + z = 1 \implies z = y$ From (2): $1 + y - z = 1 \implies y = z$ Both equations give the same relationship, $y = z$.

Now substitute $x=1$ and $z=y$ into equation (3): $1 - 3y + 3(y) = 2$ $1 - 3y + 3y = 2$ $1 = 2$

Step 2: Interpret the result. The result $1 = 2$ is a contradiction. This means the system of equations is inconsistent and has no solution. Therefore, the three planes do not have a common point of intersection.

Conclusion for Statement 2: STATEMENT - 2 is True.

Analysis of Statement 1: At least two of the lines ${L_1},{L_2}$ and ${L_3}$ are non-parallel.

The direction vector of the line of intersection of two planes is given by the cross product of their normal vectors.

Step 1: Find the direction vector of line L1. ${L_1}$ is the intersection of ${P_2}$ and ${P_3}$. Its direction vector $\vec{d_1}$ is parallel to $\vec{n_2} \times \vec{n_3}$. $\vec{d_1} = \vec{n_2} \times \vec{n_3} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 1 & -3 & 3 \end{vmatrix} = \mathbf{i}(3-3) - \mathbf{j}(3 - (-1)) + \mathbf{k}(-3 - 1) = 0\mathbf{i} - 4\mathbf{j} - 4\mathbf{k} = \langle 0, -4, -4 \rangle$ So, $\vec{d_1}$ is parallel to $\langle 0, 1, 1 \rangle$.

Step 2: Find the direction vector of line L2. ${L_2}$ is the intersection of ${P_3}$ and ${P_1}$. Its direction vector $\vec{d_2}$ is parallel to $\vec{n_3} \times \vec{n_1}$. $\vec{d_2} = \vec{n_3} \times \vec{n_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 3 \\ 1 & -1 & 1 \end{vmatrix} = \mathbf{i}(-3 - (-3)) - \mathbf{j}(1 - 3) + \mathbf{k}(-1 - (-3)) = 0\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} = \langle 0, 2, 2 \rangle$ So, $\vec{d_2}$ is parallel to $\langle 0, 1, 1 \rangle$.

Step 3: Find the direction vector of line L3. ${L_3}$ is the intersection of ${P_1}$ and ${P_2}$. Its direction vector $\vec{d_3}$ is parallel to $\vec{n_1} \times \vec{n_2}$. $\vec{d_3} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \mathbf{i}(1-1) - \mathbf{j}(-1 - 1) + \mathbf{k}(1 - (-1)) = 0\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} = \langle 0, 2, 2 \rangle$ So, $\vec{d_3}$ is parallel to $\langle 0, 1, 1 \rangle$.

Step 4: Compare the direction vectors. We have found that $\vec{d_1} \parallel \langle 0, 1, 1 \rangle$, $\vec{d_2} \parallel \langle 0, 1, 1 \rangle$, and $\vec{d_3} \parallel \langle 0, 1, 1 \rangle$. This means that all three lines of intersection, ${L_1}, {L_2}, {L_3}$, are parallel to each other.

Conclusion for Statement 1: The statement says "At least two of the lines ${L_1},{L_2}$ and ${L_3}$ are non-parallel". Since all three lines are parallel, this statement is False.

Final Evaluation

• STATEMENT - 1 is False.
• STATEMENT - 2 is True.

This corresponds to option D.

Geometrically, since the three planes do not have a common point and their lines of intersection are parallel, they form a triangular prism.