\(\displaystyle{3}{x}-{2}{y}+{z}={1}\)

\(\displaystyle{2}{x}+{y}-{3}{z}={3}\)

FInd a point of intersection. Set z=0, solve for x and y.

\(\displaystyle{3}{x}-{2}{y}+{0}={1}\to{x}={\frac{{{1}}}{{{3}}}}{\left({2}{y}+{1}\right)}\)

\(\displaystyle{2}{x}+{y}-{0}={3}\)

\(\displaystyle{2}\cdot{\frac{{{1}}}{{{3}}}}{\left({2}{y}+{1}\right)}+{y}={3}\)

\(\displaystyle{2}{\left({2}{y}+{1}\right)}+{3}{y}={9}\)

\(\displaystyle{7}{y}+{2}={9}\)

\(\displaystyle{7}{y}={7}\)

\(\displaystyle{y}={1}\)

\(\displaystyle{x}={\frac{{{1}}}{{{3}}}}{\left({2}{\left({1}\right)}+{1}\right)}={1}\)

Point 1: \(\displaystyle{\left({x}_{{0}},{y}_{{0}},{z}_{{0}}\right)}={\left({1},{1},{0}\right)}\)

We can use the cross product to find a direction vector for the line.

The cross product of the normals to the planes results in a vector that is orthogonal to both, and would be parallel to the line of intersection.

\(\displaystyle{<}{a},{b},{c}{>}\times{<}{d},{e},{f}\ge{<}{b}{f}-{c}{e},{c}{d}-{a}{f},{a}{e}-{b}{d}{>}\)

\(\displaystyle{<}{3},-{2},{1}{>}\times{<}{2},{1},-{3}\ge{<}-{2}{\left(-{3}\right)}-{1}{\left({1}\right)},{1}{\left({2}\right)}-{3}{\left(-{3}\right)},{3}{\left({1}\right)}-{\left(-{2}\right)}{\left({2}\right)}{>}\)

\(\displaystyle={<}{5},{11},{7}{>}\)

Plug the point into \(\displaystyle{\left({x}_{{0}},{y}_{{0}},{z}_{{0}}\right)}\) and the vector found into \(\displaystyle{<}{a},{b},{c}{>}\)

\(\displaystyle{x}={x}_{{0}}+{a}{t}\)

\(\displaystyle{y}={y}_{{0}}+{b}{t}\)

\(\displaystyle{z}={z}_{{0}}+{c}{t}\)

\(\displaystyle{x}={1}+{\left({5}\right)}{t}\)

\(\displaystyle{y}={1}+{\left({11}\right)}{t}\)

\(\displaystyle{z}={0}+{\left({7}\right)}{t}\)

\(\displaystyle{x}={1}+{5}{t}\)

\(\displaystyle{y}={1}+{11}{t}\)

\(\displaystyle{z}={7}{t}\)

Result: \(\displaystyle{x}={1}+{5}{t}\ {y}={1}+{11}{t}\ {z}={7}{t}\)

\(\displaystyle{2}{x}+{y}-{3}{z}={3}\)

FInd a point of intersection. Set z=0, solve for x and y.

\(\displaystyle{3}{x}-{2}{y}+{0}={1}\to{x}={\frac{{{1}}}{{{3}}}}{\left({2}{y}+{1}\right)}\)

\(\displaystyle{2}{x}+{y}-{0}={3}\)

\(\displaystyle{2}\cdot{\frac{{{1}}}{{{3}}}}{\left({2}{y}+{1}\right)}+{y}={3}\)

\(\displaystyle{2}{\left({2}{y}+{1}\right)}+{3}{y}={9}\)

\(\displaystyle{7}{y}+{2}={9}\)

\(\displaystyle{7}{y}={7}\)

\(\displaystyle{y}={1}\)

\(\displaystyle{x}={\frac{{{1}}}{{{3}}}}{\left({2}{\left({1}\right)}+{1}\right)}={1}\)

Point 1: \(\displaystyle{\left({x}_{{0}},{y}_{{0}},{z}_{{0}}\right)}={\left({1},{1},{0}\right)}\)

We can use the cross product to find a direction vector for the line.

The cross product of the normals to the planes results in a vector that is orthogonal to both, and would be parallel to the line of intersection.

\(\displaystyle{<}{a},{b},{c}{>}\times{<}{d},{e},{f}\ge{<}{b}{f}-{c}{e},{c}{d}-{a}{f},{a}{e}-{b}{d}{>}\)

\(\displaystyle{<}{3},-{2},{1}{>}\times{<}{2},{1},-{3}\ge{<}-{2}{\left(-{3}\right)}-{1}{\left({1}\right)},{1}{\left({2}\right)}-{3}{\left(-{3}\right)},{3}{\left({1}\right)}-{\left(-{2}\right)}{\left({2}\right)}{>}\)

\(\displaystyle={<}{5},{11},{7}{>}\)

Plug the point into \(\displaystyle{\left({x}_{{0}},{y}_{{0}},{z}_{{0}}\right)}\) and the vector found into \(\displaystyle{<}{a},{b},{c}{>}\)

\(\displaystyle{x}={x}_{{0}}+{a}{t}\)

\(\displaystyle{y}={y}_{{0}}+{b}{t}\)

\(\displaystyle{z}={z}_{{0}}+{c}{t}\)

\(\displaystyle{x}={1}+{\left({5}\right)}{t}\)

\(\displaystyle{y}={1}+{\left({11}\right)}{t}\)

\(\displaystyle{z}={0}+{\left({7}\right)}{t}\)

\(\displaystyle{x}={1}+{5}{t}\)

\(\displaystyle{y}={1}+{11}{t}\)

\(\displaystyle{z}={7}{t}\)

Result: \(\displaystyle{x}={1}+{5}{t}\ {y}={1}+{11}{t}\ {z}={7}{t}\)