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Find all points on the intersection of three given planes

Consider three planes given by the Cartesian equations

    \[ 3x + y + z = 5, \qquad 3x + y + 5z = 7, \qquad x - y + 3z = 3. \]

Find all of the points which are on the intersections of these three planes.


The points on the intersection must satisfy the system of equations

    \begin{align*}  3x + y + z &= 5, \\  3x + y + 5z &= 7, \\  x - y + 3z &= 3. \end{align*}

To find the points satisfying these equations we row-reduce the coefficient matrix of the system (if you don’t have any linear algebra, then you can use Gaussian elimination the long way),

    \begin{align*}  \begin{pmatrix*}[r] 3 & 1 & 1 & 5 \\ 3 & 1 & 5 & 7 \\ 1 & -1 & 3 & 3 \end{pmatrix*} &\rightarrow   \begin{pmatrix*}[r] 0 & 4 & -8 & -4 \\ 0 & 4 & -4 & -2 \\ 1 & -1 & 3 & 3 \end{pmatrix*} \\[9pt]  &\rightarrow \begin{pmatrix*}[r] 1 & -1 & 3 & 3 \\ 0 & 1 & -2 & -1 \\ 0 & 2 & -2 & -1 \end{pmatrix*} \\[9pt]  &\rightarrow \begin{pmatrix*}[r] 1 & 0 & 1 & 2 \\ 0 & 1 & -2 & -1 \\ 0 & 0 & 2 & 1 \end{pmatrix*} \\[9pt]  & \rightarrow \begin{pmatrix*}[r] 1 & 0 & 0 & \frac{3}{2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} \end{pmatrix*}. \end{align*}

Therefore, we have x = \frac{3}{2}, y = 0, and z = \frac{1}{2}. So, the only point on the intersection is the point \left \{ \left( \frac{3}{2}, 0 \frac{1}{2} \right) \right\}.

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