Given A(0,5,0) , B(1,-2,0) and C(-6,5,0) , find the point of…

Given $A(0,5,0)$, $B(1,-2,0)$ and $C(-6,5,0)$, find the point of the plane $3x-2y+z-1=0$ that is equidistant from the sides of triangle $ABC$.

Explanation of Method

Points in space equally far from the three side lines of a triangle lie on vertical lines through the incenter and excenters.

Step 1: Distance to a side in space

The triangle lies in $z=0$. For $P(x,y,z)$, let $P'=(x,y,0)$. If $P'$ is at planar distance $\delta$ from a side line, the perpendicular from $P$ to that line has length

\[ \sqrt{\delta^2+z^2}\qquad\text{(three perpendiculars theorem).} \]

So $P$ is equidistant from the three sides if and only if $P'$ is equidistant from them in the plane $z=0$. This means $P'$ is the incenter, or one of the excenters if the sides are taken as full lines.

Step 2: Incenter

The sides are $AB:\ 7x+y-5=0$, $BC:\ x+y+1=0$ and $CA:\ y=5$. Their lengths are $a=|BC|=7\sqrt2$, $b=|CA|=6$ and $c=|AB|=5\sqrt2$.

\[ I_0=\frac{aA+bB+cC}{a+b+c}=\left(\sqrt2-3,\ 6-2\sqrt2\right),\qquad r=2\sqrt2-1. \]

Check: the distances to the three sides are $|1-2\sqrt2|$, $\tfrac{4-\sqrt2}{\sqrt2}$ and $\tfrac{20-5\sqrt2}{5\sqrt2}$. All equal $2\sqrt2-1$. $\checkmark$

Step 3: Intersect the vertical line with the plane

\[ z=1-3x+2y=1-3(\sqrt2-3)+2(6-2\sqrt2)=22-7\sqrt2 . \]

\[P\left(\sqrt2-3,\ 6-2\sqrt2,\ 22-7\sqrt2\right)\approx(-1.59,\ 3.17,\ 12.10)\]

The common distance is $\sqrt{r^2+z^2}=\sqrt{591-312\sqrt2}\approx12.24$.

$A$ $B$ $C$ $I_0$ $P$

$P$ lies on the vertical line through the incenter $I_0$ of $\triangle ABC$ (height $22-7\sqrt2\approx12.1$, drawn compressed). Its perpendiculars to the three sides (red) end at the touching points of the incircle, so they have the same length $\sqrt{r^2+z^2}$.

Result

\[\boxed{\,P\left(\sqrt2-3,\ 6-2\sqrt2,\ 22-7\sqrt2\right)\approx(-1.59,\ 3.17,\ 12.10)\,}\]

Remark

Correction: the source drops the orthogonal perpendicular from $I_0$ to the plane and gets $\left(\tfrac{24-7\sqrt2}{14},\tfrac{20-7\sqrt2}7,\tfrac{22-7\sqrt2}{14}\right)$. That point is in the plane, but its distances to the three sides are about $1.00$, $2.59$ and $3.66$, so they are not equal. Moving along the normal of the tilted plane changes the horizontal position. The point must stay directly above $I_0$.

Excenters: if "sides" means the full side lines, there are three more solutions, one above each excenter:

\[ I_A(-3-6\sqrt2,\,-4-3\sqrt2):\ \ z=2+12\sqrt2, \]

\[ I_B(-3-\sqrt2,\,6+2\sqrt2):\ \ z=22+7\sqrt2, \]

\[ I_C(-3+6\sqrt2,\,-4+3\sqrt2):\ \ z=2-12\sqrt2 . \]

If "sides" means the segments, only the incenter solution remains.