Prove that if in a tetrahedron if two pairs of opposite edges are perpendicular then the third pair is also perpendicular. 
Prove that if in a tetrahedron if two pairs of opposite edges are
  perpendicular then the third pair is also perpendicular.

Method: 
$\vec a+ \vec b  + \vec c + \vec d +\vec e + \vec f = 0 \tag0$ where the vectors represent the sides. 
also, $(\vec a + \vec b). (\vec c+\vec d)= 0 \tag{1}$
$(\vec e+\vec f) .(\vec c+\vec d)= 0 \tag{2}$
From 1 and 2, 
$\vec e + \vec f = k(\vec a+ \vec b)$
Substituting in $0$,
$\vec a + \vec b = -\dfrac{\vec c + \vec d}{k+1}$
Now, $(\vec e+ \vec f). (\vec a+\vec b) = -\dfrac 1{k+1}((\vec a + \vec b). (\vec c+\vec d))= 0$ 
Hence, the third pair too is perpendicular. 
QED. 
Is my method to solve it correct? 
 A: I don't think that your way is true because it's not clear how you define $\vec{a},$ $\vec{b},$ $\vec{c}$, $\vec{d}$, $\vec{e}$ and $\vec{f}$ and why $$\vec{a}+\vec{b}+\vec{c}+\vec{d}+\vec{e}+\vec{f}=\vec{0}?$$
I like the following way.
Let $ABCD$ our tetrahedron such that $AB\perp DC$ and $AC\perp BD$.
We need to prove that $AD\perp BC$.
Indeed, let $\vec{DA}=\vec{a},$ $\vec{DB}=\vec{b}$ and $\vec{DC}=\vec{c}.$
Thus, $$\vec{c}\cdot(\vec{a}-\vec{b})=0$$ and
$$\vec{b}\cdot(\vec{a}-\vec{c})=0,$$ which gives
$$\vec{a}\cdot\vec{c}=\vec{b}\cdot\vec{c}$$ and
$$\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c},$$ which gives
$$\vec{a}\cdot\vec{c}=\vec{a}\cdot\vec{b}$$ or
$$\vec{a}\cdot(\vec{b}-\vec{c})=0$$ or
$$AD\perp BC.$$
A: In a tetrahedron $\,ABCD,\,$ perpendicular project the vertex $\,D\,$ onto the triangular face $\,ABC\,$ and call it $\,O.\,$
Edge $\,AB\,$ is perpendicular to $\,CD,\,$ if and only if $\,AB\,$ is perpendicular to $\,CO\,$ which is the altitude of vertex $\,C\,$ to opposite side $\,AB.\,$ Similarly for edge $\,AC\,$
perpendicular to $\,BD\,$ and thus the two altitudes meet at point $\,O.\,$ The three altitudes of triangle $\,ABC\,$ meet at the orthocenter $\,O.\,$ Therefore the third pair of edges 
$\,BC\,$ and $\,AD\,$ are also perpendicular.
Notice that in this proof we used the special case where $\,ABCD\,$ is a quadrilateral in a plane. If $\,D\,$ is interior to triangle $\,ABC\,$ then it is the orthocenter of triangle $\,ABC.\,$ In general, each of the four vertices is the orthocenter of the triangle formed by the other three vertices. The result for a tetrahedron $\,ABCD\,$ is a natural generalization of the planar quadrilateral orthocenter version.
A: Let $OABC$ be a tetrahedron, and define $a := |OA|$, $b := |OB|$, $c := |OC|$, $\alpha := \angle BOC$, $\beta := \angle COA$, $\gamma := \angle AOB$.
If $\overline{OA}\perp\overline{BC}$, then the plane through $\overline{BC}$, perpendicular to $\overline{OA}$, meets $\overline{OA}$ at, say, $D$. Then $\overline{BD}\perp\overline{OA}$ and $\overline{CD}\perp\overline{BC}$, so that
$$b \cos\gamma = |\overline{AD}| = c \cos\beta \qquad\to\qquad
\frac{\cos\beta}{b}=\frac{\cos\gamma}{c} \tag{1}$$
Likewise, if $\overline{OB}\perp\overline{CA}$, 
$$\frac{\cos\gamma}{c} = \frac{\cos\alpha}{a} \tag{2}$$
We conclude from $(1)$ and $(2)$ that $$\frac{\cos\alpha}{a}=\frac{\cos\beta}{b} \tag{3}$$
which implies $\overline{OC}\perp\overline{AB}$. $\square$
This is effectively @MichaelRozenberg's vector proof, without the vectors. I find it useful, as it provides a nice characterization of these triply-perpendicular  (aka, "orthocentric") tetrahedra:
$$\frac{\cos\alpha}{a} = \frac{\cos\beta}{b}=\frac{\cos\gamma}{c} \tag{$\star$}$$
A: Let the  tetrahedron $T$ be spanned from $0$ by the three vectors $a$, $b$, $c$. The three equations 
$$\eqalign{v+w&=a\cr u\!\qquad+w&=b\cr u+v\qquad &=c\cr}\tag{1}$$
 determine three linearly independent vectors $u$, $v$, $w$. From $(1)$ it then follows that the vertices of $T$ are vertices of the parallelepiped $P$ spanned by $u$, $v$, $w$, and opposite edges of $T$ are diagonals of opposite faces of $P$. Two such edges of $T$ are orthogonal iff the corresponding faces of $P$ are rhombi. If we have two pairs of orthogonal opposite edges of $T$ we can therefore conclude that $|u|=|v|=|w|$, which then implies that the third pair of opposite edges of $T$ is orthogonal as well.

