The equation of plane `pi` in the form
`ax +by+cz =d` is equivalent to the plane being given by a point in it `(x_0,y_0, z_0)` and the normal to the plane at this particular point.
`a(x-x_0) +b(y-y_0)+c(z-z_0) =0 <=> <a,b,c>*(R-R_0) =0`
Thus the vector `<a,b,c>` is a vector perpendicular to the plane `pi` .
Now, the line `L` given by the equations:
`(x-1)/5 =(y+4)/12 and z=2`
is equivalent to
`y = 12/5*x -32/5 and z=2`
A point on the line is `A=(1,-4,2) hArr x =1, y=-20/5, z=2`
and its director vector is `<5,12,0>` .
The parametric equation of the line `L` is
`(x,y,z) = (1,-4,2)+t<5,12,0> `
The condition of the line `L` intersecting the plane `pi` in a single point is equivalent to the vectors `V_1=<a,b,c>` and `V_2=<5,12,0>` not being perpendicular (since `<a,b,c>` is the normal to the plane). (Otherwise line `L` would be enclosed in the plane.)
`<a,b,c>** <5,12,0> != 0` where `**` is the dot product.
`5a +12b !=0` or equivalent `a!= -12/5*b`
Thus the condition necessary and sufficient for the line `L` to intersect the plane `pi` is a single point is `a!= -12/5*b`