Given Information
We are given with two solution sets: \[\begin{array}{l} {x_1} - 3{x_2} + 5{x_3} = 0\\ {x_1} - 3{x_2} + 5{x_3} = 4 \end{array}\] We have to compare these solutions.

Step-1: Solutions set-1
General solution of the equation is: \[{x_1} = 3{x_2} - 5{x_3}\] Write in parametric form: \[x = \left[ {\begin{array}{*{20}{c}} {3{x_2} - 5{x_3}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = {x_2}\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0 \end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}} { - 5}\\ 0\\ 1 \end{array}} \right]\] Therefore,

The solution set is a plane passing thorough the origin.

Step-2: Solutions set-2
General solution of the equation is: \[{x_1} = 4 + 3{x_2} - 5{x_3}\] Write in parametric form: \[\begin{array}{l} x = \left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {4 + 3{x_2} - 5{x_3}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 4\\ 0\\ 0 \end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0 \end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}} { - 5}\\ 0\\ 1 \end{array}} \right] \end{array}\] Therefore,

The solution set is a plane passing thorough the point $\left[ {\begin{array}{*{20}{l}} 4\\ 0\\ 0 \end{array}} \right]$

Step-3: Comparison of Solutions sets
The only difference between these solutions is that, the planes pass through different points.