Answer
$14x$ $-$ $6y$ $-$ $10z$ $=$ $9$
Work Step by Step
All points equidistant from $A$$($$-1$,$5$,$3$$)$ and $B$ $($$6$,$2$,$-2$$)$
So, we need to find a plane in three-dimensional that all the points in that plane equidistant from $A$$($$-1$,$5$,$3$$)$ and $B$ $($$6$,$2$,$-2$$)$
Let $P$ be $($ $x$ , $y$ , $z$ $)$,
$AP$ $=$ $\sqrt{(x-(-1))^{2} +(y-5)^{2}+(z-3)^{2}}$
$BP$ $=$ $=$ $\sqrt{(x-6)^{2} +(y-2)^{2}+(z-(-2))^{2}}$
$AP$ $=$ $BP$
$(x+1)^{2} +(y-5)^{2}+(z-3)^{2}$ $=$ $(x-6)^{2} +(y-2)^{2}+(z+2)^{2}$
$x^{2}$ $+$ $1$ $+$ $2x$ $+$ $y^{2}$ $+$ $25$ $-$ $10y$ $+$ $z^{2}$ $+$ $9$ $-$ $6z$ $=$ $x^{2}$ $+$ $36$ $-$ $12x$ $+$ $y^{2}$ $+$ $4$ $-$ $4y$ $+$ $z^{2}$ $+$ $4$ $+$ $4z$
$2x$ $-$ $10y$ $-$ $6z$ $+$ $35$ $=$ $-12x$ $-$ $4y$ $+$ $4z$ $+$ $44$
$14x$ $-$ $6y$ $-$ $10z$ $=$ $9$
$Therefore$, $every$ $point$ $satisfies$ $the$ $equation$ $14x$ $-$ $6y$ $-$ $10z$ $=$ $9$ equidistant from $A$$($$-1$,$5$,$3$$)$ and $B$ $($$6$,$2$,$-2$$)$
