Find the point of the plane 2x-3y-z+1=0 equidistant from…

Find the point of the plane $2x-3y-z+1=0$ equidistant from $A(2,3,0)$, $B(4,2,0)$ and $C(1,0,0)$.

Explanation of Method

The point of a plane at the same distance from three given points.

Step 1: Bisector planes

$|PA|=|PB|$ and $|PB|=|PC|$, after cancelling the squares: $(x-2)^2+(y-3)^2=(x-4)^2+(y-2)^2\ \Longrightarrow\ 4x-2y=7$.

\[ (x-4)^2+(y-2)^2=(x-1)^2+y^2\ \Longrightarrow\ 6x+4y=19. \]

Neither equation contains $z$, because $A,B,C$ all lie in $z=0$. Together they give the vertical line through the circumcenter $K$ of $\triangle ABC$.

Step 2: Solve

\[ 14x=33\ \Rightarrow\ x=\tfrac{33}{14},\quad y=\tfrac{17}{14},\quad z=2x-3y+1=\tfrac{29}{14}. \]

\[P\left(\tfrac{33}{14},\ \tfrac{17}{14},\ \tfrac{29}{14}\right),\qquad |PA|=|PB|=|PC|=\tfrac{\sqrt{1491}}{14}\approx2.76\]

Check: $|PA|^2=\tfrac{25+625+841}{196}$, $|PB|^2=\tfrac{529+121+841}{196}$ and $|PC|^2=\tfrac{361+289+841}{196}$ all equal $\tfrac{1491}{196}$. $\checkmark$

$A$ $B$ $C$ $K$ $P$

$P\left(\tfrac{33}{14},\tfrac{17}{14},\tfrac{29}{14}\right)$ lies in the blue plane, directly above the circumcenter $K$ of $\triangle ABC$, at distance $\tfrac{\sqrt{1491}}{14}\approx2.76$ from each vertex.

Result

\[\boxed{\,P\left(\tfrac{33}{14},\ \tfrac{17}{14},\ \tfrac{29}{14}\right),\qquad |PA|=|PB|=|PC|=\tfrac{\sqrt{1491}}{14}\approx2.76\,}\]

Remark

Correction: the source first wrote the bisector of $AB$ as $4x-2y-11=0$, which gave $P\left(\tfrac{41}{14},\tfrac5{14},\tfrac{81}{14}\right)$. That point fails the check. The correct plane is $4x-2y=7$, which gives the answer above (the source fixed this later).