The points equidistant from the point $O\left( {0,0,0} \right),A\left( {a,0,0} \right),B\left( {0,b,0} \right)$ and $C\left( {0,0,c} \right)$ has the coordinates.
1) $\left( {a,b,c} \right)$
2) $\left( {\dfrac{a}{2},\dfrac{b}{2},\dfrac{c}{2}} \right)$
3) $\left( {\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}} \right)$
4) $\left( {\dfrac{a}{4},\dfrac{b}{4},\dfrac{c}{4}} \right)$

Hint: First of all we will let a point $P\left( {x,y,z} \right)$ and then find the distance from it to each of the given points using the distance formula, $\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2} + {{\left( {{z_1} - {z_2}} \right)}^2}} $. Since, each point is equidistant from the point $P\left( {x,y,z} \right)$. Equate the distance $PO$ to $PA$ and calculate the value of $x$.
Similarly, we will equate $PO$ to $PB$ and calculate the value of $y$ and $PO$ to $PC$ and calculate the value of $z$. Also, we will write the final answer in coordinate form.

Complete step by step answer:

In this type of questions, where we have to find a point, we first let that point.
So, let $P\left( {x,y,z} \right)$ be a point such that the point is equidistant from the point $O\left( {0,0,0} \right),A\left( {a,0,0} \right),B\left( {0,b,0} \right)$ and $C\left( {0,0,c} \right)$.
Next, let us find the distances of all the points from $P$ using the formula, $\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2} + {{\left( {{z_1} - {z_2}} \right)}^2}} $.
We begin by finding the distance $PO$ where coordinates of $P$ are $\left( {x,y,z} \right)$ and coordinates of $O$ are $\left( {0,0,0} \right)$.
$
  PO = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2} + {{\left( {z - 0} \right)}^2}} \\
  PO = \sqrt {{x^2} + {y^2} + {z^2}} \\
 $
Similarly, find the distance $PA$ where coordinates of $P$are $\left( {x,y,z} \right)$ and coordinates of $A$ are $\left( {a,0,0} \right)$.
$
  PA = \sqrt {{{\left( {a - x} \right)}^2} + {{\left( {0 - y} \right)}^2} + {{\left( {0 - z} \right)}^2}} \\
  PA = \sqrt {{{\left( {a - x} \right)}^2} + {y^2} + {z^2}} \\
$
Next, find the distance $PB$ where coordinates of $P$are $\left( {x,y,z} \right)$ and coordinates of $B$ are $\left( {0,b,0} \right)$.
$
  PB = \sqrt {{{\left( {0 - x} \right)}^2} + {{\left( {b - y} \right)}^2} + {{\left( {0 - z} \right)}^2}} \\
  PB = \sqrt {{x^2} + {{\left( {b - y} \right)}^2} + {z^2}} \\
 $
And lastly, find the distance $PC$ where coordinates of $P$are $\left( {x,y,z} \right)$ and coordinates of $C$ are $\left( {0,0,c} \right)$.
$
  PC = \sqrt {{{\left( {0 - x} \right)}^2} + {{\left( {0 - y} \right)}^2} + {{\left( {c - z} \right)}^2}} \\
  PC = \sqrt {{x^2} + {y^2} + {{\left( {c - z} \right)}^2}} \\
 $
Since, each point is equidistant, let $PO = PA$
$\sqrt {{x^2} + {y^2} + {z^2}} = \sqrt {{{\left( {a - x} \right)}^2} + {y^2} + {z^2}} $
Squaring both sides, we get,
\[{x^2} + {y^2} + {z^2} = {\left( {a - x} \right)^2} + {y^2} + {z^2}\]
Terms \[{y^2}\] and \[{z^2}\] gets cancelled and we get,
$
  {x^2} = {\left( {a - x} \right)^2} \\
  x = \pm \left( {a - x} \right) \\
$
If we take, \[\left( {a - x} \right)\], then,
$
  x = a - x \\
  2x = a \\
  x = \dfrac{a}{2} \\
$
If we take, \[\left( {a - x} \right)\], then,
$
  x = - a + x \\
   - a = 0 \\
$
This does not give us the value of $x$.

Similarly, let $PO = PB$
$\sqrt {{x^2} + {y^2} + {z^2}} = \sqrt {{x^2} + {{\left( {b - y} \right)}^2} + {z^2}} $
Squaring both sides, we get,
\[{x^2} + {y^2} + {z^2} = {x^2} + {\left( {b - y} \right)^2} + {z^2}\]
Terms \[{x^2}\] and \[{z^2}\] gets cancelled and we get,
$
  {y^2} = {\left( {b - y} \right)^2} \\
  y = \pm \left( {b - y} \right) \\
$
If $y = b - y$, then, we get,
$
  y = b - y \\
  2y = b \\
  y = \dfrac{b}{2} \\
$
If we take $y = - \left( {b - y} \right)$, then,
$
  y = - \left( {b - y} \right) \\
  y = - b + y \\
  b = 0 \\
$
This case does not give any value of $y$.
Also, $PO = PC$
$\sqrt {{x^2} + {y^2} + {z^2}} = \sqrt {{x^2} + {y^2} + {{\left( {z - c} \right)}^2}} $
Squaring both sides, we get,
\[{x^2} + {y^2} + {z^2} = {x^2} + {y^2} + {\left( {z - c} \right)^2}\]
Terms \[{y^2}\] and \[{x^2}\] gets cancelled and we get,
$
  {z^2} = {\left( {c - z} \right)^2} \\
  z = \pm \left( {c - z} \right) \\
$
If we take $z = \left( {c - z} \right)$, then we get,
$
  z = c - z \\
  2z = c \\
  z = \dfrac{c}{2} \\
$
If we take $z = - \left( {c - z} \right)$, then we get,
$
  z = - \left( {c - z} \right) \\
  z = - c + z \\
  c = 0 \\
$
This case will not give the value of $z$.
Write the answer in coordinates form.
Therefore, point $P\left( {x,y,z} \right)$ is $\left( {\dfrac{a}{2},\dfrac{b}{2},\dfrac{c}{2}} \right)$.
Hence, option B is the correct option.

Note: It is important to square both sides, after equating the distances to avoid mistakes in calculation. Also, while solving the variables, there are two answers possible when square-root is done. Take the condition that holds true, that if we take \[x = - \left( {a - x} \right)\], then we will get, \[x = - a + x \Rightarrow a = 0\], which is incorrect.