<s>[INST] Volume of a parallelepiped when not given values for three vectors There is a parallelepiped determined by three dimensional vectors x, y, and z. The volume of this parallelepiped is $11$. What is the volume of the parallelepiped determined by the three dimension vectors x+2y, y+2z, z+2x? 
I tried substituting constant values for x, y, and z, but it became very messy quickly. How should I approach this problem? 
 [/INST] Hint:
Volume of original parallelepiped = $|\det(A)|$, where the columns of $A$ are $\mathbf x$, $\mathbf y$, $\mathbf z$.
Volume of new parallelepiped = $|\det(B)|$, where the columns of $B$ are $\mathbf x +2\mathbf y$, $\mathbf y+2\mathbf z$, $\mathbf z+2\mathbf x$.
Alternative Hint:
Volume of original parallelepiped = $|\mathbf x \cdot (\mathbf y \times \mathbf z)|$.
Volume of new parallelepiped = $|(\mathbf x +2\mathbf y) \cdot [(\mathbf y+2\mathbf z) \times (\mathbf z+2\mathbf x)]|$.

EDIT:  It's been long enough that I'd like to just write the answer down.
Determinant Approach
If we consider a determinant not as a function on square matrices, but as a function on the columns (or rows) of square matrices, then we can see that $\det(\mathbf x, \mathbf y, \mathbf z) = 11$ from the problem and the above hint.
The $\det$ function is multilinear and antisymmetric.  Thus $$\begin{align}\det(\mathbf x +2\mathbf y, \mathbf y+2\mathbf z, \mathbf z+2\mathbf x) &= \det(\mathbf x,\mathbf y+2\mathbf z, \mathbf z+2\mathbf x) + 2\det(\mathbf y, \mathbf y+2\mathbf z, \mathbf z+2\mathbf x) \\ &= [\det(\mathbf x,\mathbf y, \mathbf z+2\mathbf x) + 2\det(\mathbf x,\mathbf z, \mathbf z+2\mathbf x)] \\ &\ \ \ \  + [2\det(\mathbf y, \mathbf y, \mathbf z+2\mathbf x) + 4\det(\mathbf y, \mathbf z, \mathbf z+2\mathbf x)] \\ &= \det(\mathbf x,\mathbf y, \mathbf z) + 2\det(\mathbf x,\mathbf y, \mathbf x) + 2\det(\mathbf x,\mathbf z, \mathbf z) \\ &\ \ \ \ + 4\det(\mathbf x,\mathbf z, \mathbf x) + 2\det(\mathbf y, \mathbf y, \mathbf z) + 4\det(\mathbf y, \mathbf y, \mathbf x) \\ &\ \ \ \ + 4\det(\mathbf y, \mathbf z, \mathbf z) + 8\det(\mathbf y, \mathbf z, \mathbf x) \\ &= 11 + 2(0) + 2(0) + 4(0) + 2(0) + 4(0) + 4(0) + 8(11) \\ &= 99\end{align}$$
Note that this looks like a lot of work, but after you've practiced it a few times you'll be able to immediately look at an expression like the one above and mentally throw out all of the zero terms.  The way you do this is basically just look for which combinations don't give you more than $1$ of each column vector.  Starting with $(\mathbf x +2\mathbf y, \mathbf y+2\mathbf z, \mathbf z+2\mathbf x)$, the only non-null combos are going to be $(\mathbf x, \mathbf y, \mathbf z)$ and $(2\mathbf y, 2\mathbf z, 2\mathbf x)$ as those are the only choices without a repeated $\mathbf x$, $\mathbf y$, or $\mathbf z$.
Triple Scalar Product Approach
We know that $|\mathbf x \cdot (\mathbf y \times \mathbf z)| = 11$, so then we just have to reduce the following expression to something in terms of this one:
$$\begin{align}(\mathbf x +2\mathbf y) \cdot \color{red}{[}(\mathbf y+2\mathbf z) \times (\mathbf z+2\mathbf x)\color{red}{]} &= \mathbf x \cdot \color{red}{[}(\mathbf y+2\mathbf z) \times (\mathbf z+2\mathbf x)\color{red}{]} + 2\mathbf y \cdot \color{red}{[}(\mathbf y+2\mathbf z) \times (\mathbf z+2\mathbf x)\color{red}{]} \\ &= \mathbf x \cdot\color{red}{[}\mathbf y \times (\mathbf z+2\mathbf x) +2\mathbf z \times (\mathbf z+2\mathbf x)\color{red}{]} \\ &\ \ \ \ + 2\mathbf y \cdot \color{red}{[}\mathbf y \times (\mathbf z+2\mathbf x) +2\mathbf z \times (\mathbf z + 2\mathbf x)\color{red}{]} \\ &= \mathbf x \cdot \color{red}{[}\mathbf y \times (\mathbf z+2\mathbf x)\color{red}{]} +2\mathbf x \cdot \color{red}{[}\mathbf z \times (\mathbf z+2\mathbf x)\color{red}{]} \\ &\ \ \ \ + 2\mathbf y \cdot \color{red}{[}\mathbf y \times (\mathbf z+2\mathbf x)\color{red}{]} +4\mathbf y \cdot\color{red}{[}\mathbf z \times (\mathbf z + 2\mathbf x)\color{red}{]} \\ &= \mathbf x \cdot (\mathbf y \times \mathbf z) + 2\mathbf x \cdot (\mathbf y \times \mathbf x) + 2\mathbf x \cdot (\mathbf z \times \mathbf z) + 4\mathbf x \cdot (\mathbf z \times \mathbf x) \\ &\ \ \ \ + 2\mathbf y \cdot (\mathbf y \times \mathbf z) + 4\mathbf y \cdot (\mathbf y \times \mathbf x) \\ &\ \ \ \ +4\mathbf y \cdot (\mathbf z \times \mathbf z) + 8\mathbf y \cdot (\mathbf z \times \mathbf x) \\ &= \mathbf x \cdot (\mathbf y \times \mathbf z) + 0 + 0 + 0 +0 + 0 + 0 +8 \mathbf x \cdot (\mathbf y \times \mathbf z) \\ &= 99\end{align}$$
Here we've made use of such properties as the distributivity of the cross and dot products, the fact that scalar multiplication can basically be done in whichever order (for example, in the case of the dot product: $\alpha (\mathbf a \cdot \mathbf b) = (\alpha \mathbf a) \cdot \mathbf b = \mathbf a \cdot (\alpha \mathbf b)$), and the circular shift property of the triple scalar product.
And again, some of these steps could have been done mentally by someone proficient in the application of these products.
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