To find a locus of a point P inside triangle ABC ABC is a triangle. P is a point inside ABC such that its distances from the sides of Triangle ABC are x, y, z. If a, b, c and k are given set of constants, prove that the locus of P such that ax+by+cz=k is either an empty set or a line segment or coincides with the sets of all points in triangle ABC
Now, for the locus of point P to set of all points in triangle, the constants would have have to sides of the triangle and k would have to be twice the area of triangle ABC. 
I can't figure anything out for the locus to be straight line however and that lead me to thinking ' What if I shouldn't be relating the distances to the area of the triangle' 
One other approach I tried was to assume that locus coninciding with all points would only apply for equilateral triangles, for locus to be a straight line the triangle would have to be a isoceles and for locus to be an empty set, the triangle would be scalene one.
It would be incredibly helpful if someone could point whether I should relate the various cases of the locus to types of triangles or to given constants or to something else.
 A: If $\alpha$, $\beta$, $\gamma$ are the lengths of the sides, and $s$ is double the area of the triangle, then
$$
\alpha x+\beta y+\gamma z=s,
\quad\text{that is:}\quad
z={s-\alpha x-\beta y\over\gamma}.
$$
Substitute that into $ax+by+cz=k$ and you'll get (in general) a linear equation in $x$, $y$, which can be shown to represent a line.
A: First, we simplify the problem by allowing for signed distances, and consider points outside the triangle.
We will show that the set of points that satisfy $ax+by+cz=k$ is either the null set, a line or the entire plane. Then, when restricted to the (interior) of the triangle, this becomes a null set, a line segment, or the entire interior, so the result follows.
Hint: For a point $P$, if the side lengths of the triangle are $l_a, l_b, l_c$, then $l_a x + l_b y + l_c z = 2 \Delta $.
Hint: Hence, it follows that $ax+by+cz=k$ is the entire plane iff $a:b:c:k = l_a : l_b: l_c : 2 \Delta$.
If $a:b:c = l_a : l_b: l_c $, then we either get the entire plane or the null set.
Henceforth, assume $a:b:c \neq l_a : l_b: l_c $.
Hint: For fixed values of $a, b, c, k$, if $P_1$ and $P_2$ are distinct points that satisfy the equation, then so does the entire line segment $P_1 P_2$.
Corollary: The solution set is a (affine) sub-space. It remains to show that the solution set cannot be a single point. 
Hint: For fixed values of $a, b, c, k$, if there is (at least) 1 point that satisfies it, then there are at least 2 distinct points that satisfy it. 
Suppose we have the solution $P_1 = (x_1, y_1, z_1)$, then the point $P_2 = (x_1 + bl_c - cl_b, y_1 +cl_a - al_c, z_1 + al_b-bl_a)$ will also satisfy both equations (expand and cancel terms). These are distinct points if $a:b:c \neq l_a : l_b: l_c $.
