 Solution

Using the Levi-Civita symbol and Einstein summation convention, the scalar triple product isInterchange the dummy indices $i,j$. Antisymmetry gives $S=-a_j{\epsilon }_{ijk}{a}_i{b}_k=-S$, since $a_i{a}_j=a_j{a}_i$. Over the real numbers this proves

The normal vector to the first plane and the specified positive direction give $a=(1,1,1)$. Then $a\cdot m=3$ and $a\times m=(1,1,-2)$, soNext $a\times b=(-9,9,0)$. 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 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 $f=sm=(2s,0,s)$. 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, areTherefore 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.