= Solution
Using the <Levi-Civita symbol> and <Einstein summation convention>, the <scalar triple product> is
$$
S=a_i\varepsilon_{ijk}a_jb_k.
$$
Interchange the dummy indices $i,j$. Antisymmetry gives $S=-a_j\varepsilon_{ijk}a_ib_k=-S$, since $a_ia_j=a_ja_i$. Over the <real numbers> this proves
$$
\boxed{a\cdot(a\times b)=0.}
$$
The normal <vector> to the first plane and the specified positive direction give $a=(1,1,1)$. Then $a\cdot m=3$ and $a\times m=(1,1,-2)$, so
$$
b=(3,3,-6),\qquad a+b=(4,4,-5).
$$
Next $a\times b=(-9,9,0)$. A <vector> perpendicular to both $a$ and $b$ is a scalar multiple of their <cross product>; the required length gives the two possibilities $c=(-2,2,0)$ and $c=(2,-2,0)$. The original PDF specifies approach to the \b[x-axis]. From $(4,4,-5)$ the original squared distance to that axis is $4^2+(-5)^2=41$. The two candidate endpoint squared distances are $6^2+(-5)^2=61$ and $2^2+(-5)^2=29$. Hence $c=(2,-2,0)$.
Adding the given $d$ yields $(7,0,-3)$. By the proved <scalar triple product> identity, $a\cdot b=3a\cdot(a\times m)=0$, so the only <vector> of the specified length for $e$ is the <zero vector>. Finally set $f=s\,m=(2s,0,s)$. The last point is $(7+2s,0,-3+s)$, and its plane condition becomes $2+8s=10$. Thus $s=1$ and $f=(2,0,1)$.
All waypoints, in the given distance units, are
$$
\begin{array}{c|c}
\text{stage}&\text{position}\\\hline
\text{start}&(0,0,0)\\
a&(1,1,1)\\
b&(4,4,-5)\\
c&(6,2,-5)\\
d&(7,0,-3)\\
e&(7,0,-3)\\
f&(9,0,-2)
\end{array}
$$
Therefore \b[the final location is $(9,0,-2)$]. The converted TeX omits two <vector> instructions and changes the axis; the computation above follows the original PDF.
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