The coordinates of the point P which is equid | vedclass

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(A) Let the coordinates of point $P$ be $(x, y, z)$ such that $PA = PB = PC = PO$.From $PO^2 = PA^2$,we have:$x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2$

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$\left( \frac{a}{2}, \frac{b}{2}, \frac{c}{2} \right)$

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(A) Let the coordinates of point $P$ be $(x, y, z)$ such that $PA = PB = PC = PO$.From $PO^2 = PA^2$,we have:$x^2 + y^2 + z^2 = (x - a)^2 + y^2 + z^2$$x^2 = x^2 - 2ax + a^2$$2ax = a^2 \implies x = \frac{a}{2}$ (since $a 
eq 0$).Similarly,from $PO^2 = PB^2$,we have:$x^2 + y^2 + z^2 = x^2 + (y - b)^2 + z^2$$y^2 = y^2 - 2by + b^2$$2by = b^2 \implies y = \frac{b}{2}$ (since $b 
eq 0$).Similarly,from $PO^2 = PC^2$,we have:$x^2 + y^2 + z^2 = x^2 + y^2 + (z - c)^2$$z^2 = z^2 - 2cz + c^2$$2cz = c^2 \implies z = \frac{c}{2}$ (since $c 
eq 0$).Therefore,the coordinates of point $P$ are $\left( \frac{a}{2}, \frac{b}{2}, \frac{c}{2} ight)$.

The coordinates of the point $P$ which is equidistant from the points $A(a, 0, 0)$,$B(0, b, 0)$,$C(0, 0, c)$,and $O(0, 0, 0)$ are .......... where $a, b, c \neq 0$.

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