If the planes $x - y + z + 1 = 0, \lambda x + 3y + 2z - 3 = 0, 3x + \lambda y + z - 2 = 0$ form a triangular prism then $\lambda$ is __________.
Step-by-Step Solution
Key Concept: A triangular prism is formed when three planes satisfy the condition that their normal vectors' determinant equals zero (planes meet at a line or are specially configured), which is found by setting $\begin{vmatrix} 1 & -1 & 1 \\ \lambda & 3 & 2 \\ 3 & \lambda & 1 \end{vmatrix} = 0$.
For three planes to form a triangular prism, two of them must be parallel (having the same normal vector direction) and the third must intersect both. Planes are parallel when their normal vectors are proportional. Let the normals be $\vec{n_1} = (1, -1, 1)$, $\vec{n_2} = (\lambda, 3, 2)$, and $\vec{n_3} = (3, \lambda, 1)$. For a prism, we need two parallel planes, so either $\vec{n_2} \parallel \vec{n_3}$, giving $\frac{\lambda}{3} = \frac{3}{\lambda} = \frac{2}{1}$, which yields $\lambda = 3$ and $\lambda = \frac{3}{2}$ (contradiction). Alternatively, check if $\vec{n_1} \parallel \vec{n_2}$ or $\vec{n_1} \parallel \vec{n_3}$: for $\vec{n_1} \parallel \vec{n_2}$: $\frac{1}{\lambda} = \frac{-1}{3} = \frac{1}{2}$ (impossible). For $\vec{n_1} \parallel \vec{n_3}$: $\frac{1}{3} = \frac{-1}{\lambda} = \frac{1}{1}$ (impossible). Therefore, check when the three planes don't form a proper triangular prism by requiring the determinant condition: $\begin{vmatrix} 1 & -1 & 1 \\ \lambda & 3 & 2 \\ 3 & \lambda & 1 \end{vmatrix} = 0$, which gives $\lambda = 5$ or $\lambda = -1$.
Correct Answer: 5,-1