In the fig. 

 If the coordinates of  $P$ are $(a,b,c)$ then the reflections or images 

of $P$ in $XY, YZ and ZX$- planes are 

The reflection or image of $P$$(a, b, c)$in $xy$- plane will 

be as much below the $xy$-plane as point $P$ is above it, that is,

if $P^{\mathrm{\prime }}$ is the reflection of  $P$in $xy$-plane, then $P^{\mathrm{\prime }}D=PD=c$ and $P^{\mathrm{\prime }}D$

is parallel to $OZ\text{'}$.  So, the coordinates of $P^{\mathrm{\prime }}\text{ are }(a,b,-c)$.

The image of $P(a,b,-c)$  in $yz$-plane will be as much on the back 
side of $yz$-plane as the point $P$ is on its front side. Thus, if $P$'' is

the image of $P$in $yz$-plane, then P" lies  on $PE$ such that 

$PE =EP".$

But, $PE= OA =a$. .So, the coordinates of $P^{\prime \prime }$are $(-a,b,c)$

The image of $P (a, b, c)$ in $zx-$plane  will be as much as on

 the left side of $xz$- plane as the point $P$ is on its right side. 

Thus, if $p"\text{'}$  lies on  $PF$  produced such that $PF=F{P}^{\prime \prime \mathrm{\prime }}.$

But $PF =OB =b$ So, the coordinates of $p"\text{'}$ are $(a,-b,c)$