In Figure, if the coordinates of P are (a, b, c), then the perpendicular distances of P from the coordinates axes are respectively

$\sqrt{b^2 + c^2 }, \sqrt{c^2+a^2}, \sqrt{a^2+b^2}$

PA, PB and PC are the perpendicular distances of point P from OX, OY and OZ respectively.

In right angled triangle ADP, we have

$AP^2 = AD^2 +DP^2 $

$⇒ PA = \sqrt{AD^2 +DP^2 }$

$⇒ PA = \sqrt{b^2 +c^2} $             $[∵ AD = OB = b $ and $PD = OC = c]$

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

$PB^2 = BD^2 +PD^2 $

$⇒PB = \sqrt{BD^2 +PD^2} $

$⇒PB = \sqrt{a^2 +c^2} $                 $[∵BD = OA = a $ and $ PD = OC = c]$

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

$PC^2 = PF^2 +CF^2 $

$⇒ PC = \sqrt{PF^2+CF^2}$

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

$⇒ PC = \sqrt{a^2+b^2}$