vec a=\1,2α,1\ , vec b=\2,α,α\ , vec c=\3α,2,-α\ are edges…
$\vec a=\{1,2\alpha,1\}$, $\vec b=\{2,\alpha,\alpha\}$, $\vec c=\{3\alpha,2,-\alpha\}$ are edges of a tetrahedron. $1^\circ$ Find its volume. $2^\circ$ Find $\alpha$ for which $\vec a,\vec b,\vec c$ are coplanar, and then decompose $\vec a$ along $\vec b$ and $\vec c$.
Explanation of Method
Volume of a tetrahedron with a parameter and the coplanarity condition.
Step 1: $1^\circ$ Triple product and volume
\[ [\vec a,\vec b,\vec c]=\begin{vmatrix}1&2\alpha&1\\2&\alpha&\alpha\\3\alpha&2&-\alpha\end{vmatrix}=(-\alpha^2-2\alpha)-2\alpha(-2\alpha-3\alpha^2)+(4-3\alpha^2)=6\alpha^3-2\alpha+4. \]
\[V=\frac16\,|6\alpha^3-2\alpha+4|=\left|\alpha^3-\frac{\alpha}{3}+\frac23\right|\]
(Check: $\alpha=0$ gives a determinant of $4$, $\alpha=1$ gives $8$.)
Step 2: $2^\circ$ Coplanarity
\[ 3\alpha^3-\alpha+2=(\alpha+1)(3\alpha^2-3\alpha+2)=0,\qquad \Delta=9-24\lt0, \] so the only real root is
\[\alpha=-1\]
- $\tfrac16[\vec a,\vec b,\vec c]=\alpha^3-\tfrac\alpha3+\tfrac23$
- $V(\alpha)=\left|\alpha^3-\tfrac\alpha3+\tfrac23\right|$
Step 3: Decomposition for $\alpha=-1$
\[ \vec a=\{1,-2,1\},\quad \vec b=\{2,-1,-1\},\quad \vec c=\{-3,2,1\},\qquad \vec a=\beta\vec b+\gamma\vec c: \]
\[ 2\beta-3\gamma=1,\quad -\beta+2\gamma=-2,\quad -\beta+\gamma=1\ \Longrightarrow\ \gamma=-3,\ \beta=-4. \]
\[\vec a=-4\vec b-3\vec c\]
Check: $\{-8,4,4\}+\{9,-6,-3\}=\{1,-2,1\}$ $\checkmark$
Result
\[\boxed{\,V=\frac16\,|6\alpha^3-2\alpha+4|=\left|\alpha^3-\frac{\alpha}{3}+\frac23\right|\,}\]
\[\boxed{\,\alpha=-1\,}\]
\[\boxed{\,\vec a=-4\vec b-3\vec c\,}\]