At what point does the line #(x+4)/2 = (y-6)/-1 = (z+2)/4# meet the coordinate planes?

While an untraditional definition of a line, we can try and break down what it is meaning.

In order to really define this line, I'm going to introduce a new variable #t#. #t# is equal to each of the things above, i.e.
#t = (x+4)/2 #, #t = (y-6)/-1#, #t = (z+2)/4#. If you were to solve for (x,y,z) it is clear how this would be a single parameter to define a 3D object, i.e. a 1D curve. In this case, since each of these relationships is linear, the whole function will be linear and it will be a line in 3D. Therefore, there is only one crossing of the coordinate plane.

The "coordinate plane" refers to the traditional x-y plane. This has the property that z = 0. Hence, we can plug that in to find #t#:
#(0+2)/4 = t = 1/2 #.
Therefore,
#1/2 = (x+4)/2 implies x+4 = 1 implies x = -3 #
#1/2 = (y-6)/-1 implies 2y-12 = -1 implies y = 11/2#

Therefore, it intercepts the plane at #(-3, 11/2)#.

Given equation of line:

#\frac{x+4}{2}=\frac{y-6}{-1}=\frac{z+2}{4}#

Let #\frac{x+4}{2}=\frac{y-6}{-1}=\frac{z+2}{4}=t#

#x=2t-4, \ y=-t+6# & #z=4t-2#

Thus, there is a general point on the given line #(2t-4, -t+6, 4t-2)#

Now, the line will intersect the XY-plane at the point where #z=0# hence

#4t-2=0\implies t=1/2#

hence, the point where the line intersects the XY-plane at the point

#(2(1/2)-4, -1/2+6, 4(1/2)-2)\equiv(-3, 11/2, 0)#

Now, the line will intersect the YZ-plane at the point where #x=0# hence

#2t-4=0\implies t=2#

hence, the point where the line intersects the YZ-plane at the point

#(2(2)-4, -2+6, 4(2)-2)\equiv(0, 4, 6)#

Now, the line will intersect the ZX-plane at the point where #y=0# hence

#-t+6=0\implies t=6#

hence, the point where the line intersects the ZX-plane at the point

#(2(6)-4, -6+6, 4(6)-2)\equiv(8, 0, 22)#