The volume of parallelepiped $S$ is reduced to $90\%$ of $T$.Prove that locus of $A_1$ is a plane. $T$ is a parallelepiped in which $A,B,C$ and $D$ are vertices of one face.And the face just above it has corresponding vertices $A'B'C'D'$. $T$ is now compressed to $S$ with face $ABCD$ remaining same and $A'B'C'D'$ shifted to $A_1B_1C_1D_1$ in $S$. The volume of parallelepiped $S$ is reduced to $90\%$ of $T$. Prove that locus of $A_1$ is a plane.

My Attempt:
Let $ABCD$ be the base face of the parallelepiped. And let $A'B'C'D'$ be the top face.Let $A$ be the origin and the position vectors of $B,C,D,A'$ be $\vec{b},\vec{c},\vec{d},\vec{a'}$. Then the volume of parallelepiped $T$ is $\vec{b}\cdot(\vec{d}\times\vec{a'})$
When the $T$ is compressed to $S$, then the base face remains the same and the top face becomes $A_1B_1C_1D_1$. Let the position vector of $A_1$ be $\vec{a_1}$. Then volume of $S$ is $\vec{b}\cdot(\vec{d}\times\vec{a_1})$
$S=0.9 T$
But i dont know how to do further. I am stuck here. Please help me. Thanks.
 A: The scalar triple product $[\vec u,\vec v,\vec w] = \vec u\cdot(\vec v\times\vec w)$ has the property that $[\vec u,\vec v,\vec w] = [\vec u,\vec v,\vec w +\alpha\vec u+\beta\vec v]$ for any scalars $\alpha$ and $\beta$. This corresponds to the fact that you can translate the edge of a parallelogram along its direction without changing the area. (Similarly you can translate the face of a parallelepiped in its plane without changing its volume.)
So, using this fact, you add to $\vec a_1$ any linear combination of $\vec b$ and $\vec d$ without changing the volume; this set of vectors describes a plane.
Edit: to see why the scalar triple product has this property, use the linearity of the dot and cross products.
$$\vec u\cdot(\vec v\times(\vec w+\alpha\vec u+\beta\vec v))=\vec u\cdot(\vec v\times\vec w+\alpha\vec v\times\vec u+\beta\vec v\times\vec v)$$
$$=\vec u\cdot(\vec v\times\vec w)+\alpha\vec u\cdot(\vec v\times\vec u)+\beta\vec u\cdot(\vec v\times\vec v)$$
The second term is zero because $\vec v\times\vec u$ is perpendicular to $\vec u$, and the third term is zero because $\vec v\times\vec v$ is the zero vector (as is the cross product of any vector with itself). So only the first term remains.
