In the fig. 

 if the coordinates of $P$are $(a, b, c),$then the perpendicular 

distances  of $P$from the coordinates axes are respectively 

$PA, PB and PC$  are the perpendicular distances of 

point $P$from$OX, OY and OZ$ respectively.

In right angled triangle $ADP$, we have 

$\begin{array}{l} A{P}^2=A{D}^2+D{P}^2\\ =PA=\sqrt{A{D}^2+D{P}^2}\end{array}$

$\Rightarrow PA=\sqrt{b^2+c^2} [\because AD=OB=b\text{ and }PD=OC=c]$

In right angled triangle $PDB$ right angled at $D$, we have 

$\begin{array}{l} P{B}^2=B{D}^2+P{D}^2\\ \Rightarrow PB=\sqrt{B{D}^2+P{D}^2}\end{array}$

$\Rightarrow PB=\sqrt{a^2+c^2} [\because BD=OA=a\text{ and }PD=OC=c]$

In right angled triangle $PCF$ right angled at $F$, we have 

$\begin{array}{l} P{C}^2=P{F}^2+C{F}^2\\ \Rightarrow PC=\sqrt{P{F}^2+C{F}^2}\end{array}$

$\Rightarrow PC=\sqrt{b^2+a^2} [\because PF=AD=OB=b\text{ and }CF=OA=a]$

$\Rightarrow PC=\sqrt{a^2+b^2}$