vector algebra proof I've been working on the following question
If ${\bf p}$, ${\bf q}$, and ${\bf r}$ are any vectors, demonstrate that  ${\bf a}={\bf q} + \lambda\,{\bf r}$,  ${\bf b} = {\bf r}+\mu\,{\bf p}$, and  ${\bf c} = {\bf p} + \nu\,{\bf q}$ are coplanar provided that  $\lambda\,\mu\,\nu=-1$, where $\lambda$, $\mu$, and $\nu$ are scalars. Show that this condition is satisfied when ${\bf a}$ is perpendicular to ${\bf p}$, ${\bf b}$ to ${\bf q}$, and ${\bf c}$ to ${\bf r}$.
So far I've managed to show that $\lambda\,\mu\,\nu=-1$ gives ${\bf a}={\bf q} + \lambda\,{\bf r}$,  ${\bf b} = {\bf r}+\mu\,{\bf p}$, and  ${\bf c} = {\bf p} + \nu\,{\bf q}$ are coplanar by expanding out the cross product of ${\bf a} \times {\bf b} \times {\bf c} $:
\begin{vmatrix}
{\bf (q + \lambda r) i}\ & {\bf (q + \lambda r) j} & {\bf (q + \lambda r) k}, \\
{\bf (r + \mu p) i} & {\bf (r + \mu p) j} & {\bf (r + \mu p) k} \\
{\bf (p + \nu q) i} & {\bf (p + \nu q) j} & {\bf (p + \nu q) k}
\end{vmatrix}
then the parts cancel out to give
${\bf qirjpk + qjrkpi + qkripj - qirkpj - qjripk - qkrjpi}$
+ $ \lambda \mu \nu ({\bf qirjpk + qjrkpi + qkripj - qirkpj - qjripk - qkrjpi}) $
which gives $\lambda \mu \nu = -1 $ for $ {\bf a, b, c} $ as coplanar
I'm getting stuck on showing that $\lambda \mu \nu = -1 $ where ${\bf a}$ is perpendicular to ${\bf p}$, ${\bf b}$ to ${\bf q}$, and ${\bf c}$ to ${\bf r}$.
using this perpendicular properties, I take the dot product of ${\bf a}$ and ${\bf b}$, ${\bf a}$ and ${\bf c}$, and ${\bf b}$ and ${\bf c}$:
where since ${\bf a}.{\bf p} = 0$:
${\bf a}.{\bf b} = {\bf q}.{\bf r} + {\bf \lambda r . r}$
and also since ${\bf b.q} = 0$:
${\bf a}.{\bf b} = {\bf \mu \lambda p .r} + {\bf \lambda r . r}$
equating these two and simplifying gives
${\bf \mu \lambda p = q} $
and likewise for ${\bf a.c}$ and ${\bf b.c}$ we can get
$$
{\bf \mu \nu q = r}
$$
and
$$
{\bf \lambda \nu r = p}
$$
which I think means that ${\bf p,q,r}$ are parallel and perpendicular to ${\bf a,b,c}$ which are also parallel, but I can't see how to get from here to showing that
$$\lambda \mu \nu = -1$$
any suggestions?
 A: For simplicity, I denote vectors by $p,q,...$. As far as I have understood, you want to prove the statement:
$$\forall p,q,r\in \mathbb{R}^3,\forall \lambda,\mu,\nu\in \mathbb{R}:
p\bot (q+\lambda r), q\bot (r+\mu p), r\bot (p+\nu q) \Rightarrow \lambda\mu\nu=-1.$$
However, this is not true: take any choice of pairwise orthogonal vectors $p\bot q,q\bot r,r\bot p$ then the hypothesis is satisfied regardless of the choice of $\lambda,\mu,\nu$.
Pairwise orthogonal triples $p,q,r$ are the only possible exceptions: as soon as two of the three are not orthogonal to each other, then the above implication is indeed true: suppose $r$ and $p$ are not orthogonal, that is $r\cdot p\ne 0$. Then we have $p\cdot(q+\lambda r)=q\cdot(r+\mu p)=r\cdot(p+\nu q)=0$. We obtain: $p\cdot q=-\lambda(p\cdot r)$, $q\cdot r=-\mu(q\cdot p)$, $r\cdot p=-\nu(r\cdot q)$. Using $r\cdot p=p\cdot r$ we arrive at $r\cdot p=(-\nu)(-\mu)(-\lambda)(r\cdot p)$, whence $1=(-\nu)(-\mu)(-\lambda)$.
Moreover, if $p,q,r$ are pairwise orthogonal (all $\ne 0$)  (more generally, if $p,q,r$ are linearely independent) then the condition $\lambda\mu\nu=-1$ is not only sufficient but also necessary for $q+\lambda r, r+ \mu p,p+\mu q$ to be co-planar.
