Finding the intersection points of a line with a cube The following is an old high school exercise:

Let $A = (5, 4, 6)$ and $B = (1,0,4)$ be two adjacent vertices of a cube in $\mathbb{R}^3$. The vertex $C$ lies in the $xy$-plane.
a) Compute the coordinates of the other vertices of the cube such that all $x$- and $z$-coordinates are positive.
b) Let $$g: \vec{r} = \begin{pmatrix} 10\\1\\5 \end{pmatrix} + \lambda \begin{pmatrix} 1\\1\\-1 \end{pmatrix}$$
be a line. Compute the coordinates of the intersection points of $g$ and the cube.

Question: What is the most efficient way using high school level maths to compute the intersection points by hand?
The most obvious approach would be to intersect the line with all six planes containing the cube's faces. This requires me to solve six $3 \times 3$ systems of linear equations which will probably take a while. Even worse, I will then have to check if the intersection points really belong to the surface of the cube. These are another six $3 \times 2$ systems of linear equations. Of course, the first six systems of linear equations are much easier to compute using the coordinate form of the planes, but still it will take a while.
A maybe more elegant approach would be to translate and rotate the cube such that one vertex coincides with the origin and the edges lie on the (positive) coordinate axes. Apply the same translation and rotation to $g$ and the intersection points are much easier to compute. It is also much easier to check whether or not the resulting points lie on the surface of the cube. However, in general, the rotation matrix will be very messy and therefore not advisable to do by hand.
Is there a different approach that avoids a lot of (messy) computation? I feel like there has to be some geometric property I am missing which allows for a completely different approach. After all, this is meant to be solved by high schoolers by hand.

a) finding the edges and vertices
Given
$$
A = \left( {5,4,6} \right)\quad B = \left( {1,0,4} \right)
$$
then
$$
\overline {BA}  = 6\quad \mathop {BA}\limits^ \to   = \left( {\matrix{ 4  \cr  4  \cr  2 \cr } } \right)\quad
 {\bf u} = {{\mathop {BA}\limits^ \to  } \over {\overline {BA} }}
  ={1 \over 3} \left( {\matrix{  {2}  \cr {2}  \cr {1}  \cr  } } \right)
$$
considering the conditions on $C$, we shall have
$$
\eqalign{
  & C = \left( {x,y,0} \right)\;\quad \mathop {BC}\limits^ \to   = \left( {\matrix{   {x - 1}  \cr    y  \cr    { - 4}  \cr  } } \right)  \cr 
  & \left\{ \matrix{
  \overline {BC} ^{\,2}  = 36 = \left( {x - 1} \right)^{\,2}  + y^{\,2}  + 16 \hfill \cr 
  \mathop {BA}\limits^ \to  \; \cdot \;\mathop {BC}\limits^ \to   = 0 = 4x + 4y - 12 \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{  y^{\,2}  - 2y - 8 = 0 \hfill \cr   3 - y = x \hfill \cr}  \right.\quad
 \Rightarrow \quad C = \left( {5, - 2,0} \right) \cr} 
$$
so
$$
\mathop {BC}\limits^ \to   = \left( {\matrix{   4  \cr    { - 2}  \cr    { - 4}  \cr  } } \right)\quad {\bf v}
 = {{\mathop {BC}\limits^ \to  } \over {\overline {BC} }} = {1 \over 3}\left( {\matrix{   2  \cr    { - 1}  \cr    { - 2}  \cr  } } \right)
$$
The unit vector along the third edge from $B$ will be
$$
{\bf w} = {\bf v}\; \times \;{\bf u} = {1 \over 3}\left( {\matrix{   1  \cr    { - 2}  \cr    2  \cr  } } \right)
$$
where the sign of the product is choosen to respect the condition 
for positive $x$ and $z$ coordinates.
Having the three unit vectors, it is easy to compute all the points.
$D$ will be
$$
D = C + \mathop {BA}\limits^ \to   = \left( {9,2,2} \right)
$$
while the points in the upper face will be given by
$$
A' = A + 6{\bf w} = \left( {7,0,10} \right)
$$
and similarly for the others.
b) finding the intersections with the cube
For a point $P=(x,y,z)$ to be inside the cube, the vector $\vec {BP}$ shall have its coordinates 
in the reference $(\bf u,\bf v,\bf w)$ contemporaneously within the range $[0,6]$. That is
$$
\left\{ \matrix{
  0 \le \mathop {BP\,}\limits^ \to   \cdot \;{\bf u} \le 6 \hfill \cr 
  0 \le \mathop {BP\,}\limits^ \to   \cdot \;{\bf v} \le 6 \hfill \cr 
  0 \le \mathop {BP\,}\limits^ \to   \cdot \;{\bf w} \le 6 \hfill \cr}  \right.
$$
The dot products are easily computed
$$
\eqalign{
  & \mathop {BP}\limits^ \to   = \left( {\matrix{   {10}  \cr    1  \cr    5  \cr 
 } } \right) - \left( {\matrix{   1  \cr    0  \cr    4  \cr 
 } } \right) + \lambda \left( {\matrix{   1  \cr    1  \cr    { - 1}  \cr 
 } } \right) = \left( {\matrix{   9  \cr    1  \cr    1  \cr 
 } } \right) + \lambda \left( {\matrix{   1  \cr    1  \cr    { - 1}  \cr 
 } } \right)  \cr 
  & \mathop {BP\,}\limits^ \to   \cdot \;{\bf u} = \,{1 \over 3}\left( {\matrix{   2  \cr    2  \cr    1  \cr 
 } } \right) \cdot \left( {\left( {\matrix{   9  \cr    1  \cr    1  \cr 
 } } \right) + \lambda \left( {\matrix{   1  \cr    1  \cr    { - 1}  \cr 
 } } \right)} \right) = 7 + \lambda   \cr 
  & \mathop {BP\,}\limits^ \to   \cdot \;{\bf v} = \,{1 \over 3}\left( {\matrix{   2  \cr    { - 1}  \cr    { - 2}  \cr 
 } } \right) \cdot \left( {\left( {\matrix{   9  \cr    1  \cr    1  \cr 
 } } \right) + \lambda \left( {\matrix{   1  \cr    1  \cr    { - 1}  \cr 
 } } \right)} \right) = 5 + \lambda   \cr 
  & \mathop {BP\,}\limits^ \to   \cdot \;{\bf w} = \,{1 \over 3}\left( {\matrix{   1  \cr    { - 2}  \cr    2  \cr 
 } } \right) \cdot \left( {\left( {\matrix{   9  \cr    1  \cr    1  \cr 
 } } \right) + \lambda \left( {\matrix{   1  \cr    1  \cr    { - 1}  \cr 
 } } \right)} \right) = 3 - \lambda  \cr} 
$$
and so is the system of inequalities
$$
 \left\{ \matrix{  0 \le 7 + \lambda  \le 6 \hfill \cr   0 \le 5 + \lambda  \le 6 \hfill \cr  0 \le 3 - \lambda  \le 6 \hfill \cr}  \right.\quad
   \Rightarrow \quad \left\{ \matrix{  - 7 \le \lambda  \le  - 1 \hfill \cr    - 5 \le \lambda  \le 1 \hfill \cr    - 3 \le \lambda  \le 3 \hfill \cr}  \right.\quad  
 \Rightarrow \quad  - 3 \le \lambda  \le  - 1
$$
For $\lambda$ outside of the given range, the three inequalities are not satisfied contemporaneously: the point is not inside the cube.
Thus the limits of the range are the values of $\lambda$ for which the line intersects the cube, and we have just to place them 
in the equation of the line, to get
$$
P_{\,1}  = \left( {\matrix{  10  \cr    1  \cr    5  \cr 
 } } \right) - 3\left( {\matrix{   1  \cr    1  \cr    { - 1}  \cr 
 } } \right) = \left( {\matrix{   7  \cr    { - 2}  \cr    8 \cr 
 } } \right)\quad P_{\,2}  = \left( {\matrix{   10  \cr    1  \cr    5  \cr 
 } } \right) - \left( {\matrix{   1  \cr    1  \cr    { - 1}  \cr 
 } } \right) = \left( {\matrix{  9  \cr    0  \cr    6 \cr 
 } } \right)
$$
As a rough check, note that
$$
\mathop {BP_{\,1} }\limits^ \to   = \left( {\matrix{   6  \cr    { - 2}  \cr    4  \cr 
 } } \right)\quad \mathop {BP_{\,2} }\limits^ \to   = \left( {\matrix{   8  \cr    0  \cr    2  \cr 
 } } \right)
$$
which when expressed in the $(\bf u, \bf v, \bf w)$ become
$$
\mathop {BP_{\,1} }\limits^ \to  _{\,\left( {u,v,w} \right)}  = \left( {\matrix{ 4 \cr 2 \cr 6 \cr
 } } \right)\quad \quad \mathop {BP_{\,2} }\limits^ \to  _{\;\left( {u,v,w} \right)}  = \left( {\matrix{6 \cr 42 \cr 4 \cr
 } } \right)
$$
