Consider the planes $P_1 : x - y + z = 1, P_2 : x + y - z = -1, P_3 : x - 3y + 3z = 2.$

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

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

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

Statement 1 is False, Statement 2 is True.

Vectors normal to planes $P_1, P_2 $ and $P_3$ are $\vec{n}_1 = \hat{i} - \hat{j} + \hat{k} , \hat{i} + \hat{j} - \hat{k} $ and $ \vec{n}_3 = \hat{i}- 3\hat{j} + 3\hat{k}$. Therefore, lines $L_1, L_2 $ and $L_3 $ are parallel to vectors $\vec{b}_1= \vec{n}_2 × \vec{n}_3, \vec{b}_2= \vec{n}_3 ×\vec{n}_1 $ and $ \vec{b}_3 = \vec{n}_1 × \vec{n}_2 $ respectively.

We have, $\vec{b}_1 = -4(\hat{j}+\hat{k}), \vec{b}_2 = 2 (\hat{j}+\hat{k}), \vec{b}_3= 2 (\hat{j}+\hat{k})$

Clearly, lines $L_1, L_2, L_3$ are parallel.

So, statement-1 is not true.

Consider the system of equations

$x - y + z = 1 $

$x + y - z = -1$

$x- 3y + 3z = 2 $

This system can be written as AX = B

i.e. $\begin{bmatrix}1 & -1 & 1\\1 & 1 & -1\\1 & -3 & 3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\-1\\2\end{bmatrix}$

$⇔\begin{bmatrix}1 & -1 & 1\\0 & 2 & -2\\0 & -2 & 2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\-2\\1\end{bmatrix}$ Applying $ R_2 → R_2 - R_1, R_3 → R_3 - R_1 $

$⇔\begin{bmatrix}1 & -1 & 1\\0 & 2 & -2\\0 & 0 & 0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\-2\\-1\end{bmatrix}$

Clearly, it is an inconsistent system of equations. So, given planes do not have a common point. Consequently, statement-2 is true.