Geometry & Trigonometry: Vector Line of Intersection of Two Planes

Normal vectors: $\vec{n}_1 = \langle 1, 2, -1 \rangle$ and $\vec{n}_2 = \langle 2, -1, 1 \rangle$. The direction vector $\vec{d}$ is perpendicular to both normals: $\vec{d} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & -1 & 1 \end{vmatrix} = \hat{i}(2 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - 4) = \hat{i} - 3\hat{j} - 5\hat{k}$

Set $z = 0$ in both plane equations: $x + 2y = 3$ $2x - y = 1$ Multiply second equation by 2 and add: $(x + 2y) + (4x - 2y) = 3 + 2 \implies 5x = 5 \implies x = 1$ Substitute $x = 1$: $1 + 2y = 3 \implies 2y = 2 \implies y = 1$. A common point on both planes is $P(1, 1, 0)$.

$\vec{r} = (\hat{i} + \hat{j} + 0\hat{k}) + \lambda(\hat{i} - 3\hat{j} - 5\hat{k}) = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -3 \\ -5 \end{pmatrix}, \quad \lambda \in \mathbb{R}$