Question

The distance of a point $P(a,b,c)$ from the $x$-axis is $10$ units and from the $z$ axis is $2\sqrt{41}$ units. If $a,b$ and $c$ are natural numbers, then find the value of $a+b+c$.

Answer 1

Distance of the point $P(a,b,c)$ from $x$-axis $=\sqrt{(a-a{)}^2+b^2+c^2}$
$\Rightarrow 10=\sqrt{b^2+c^2}$
$\Rightarrow 100=b^2+c^2$
$\because a,b,c\in N$
$\therefore b=6$ or $8...\left(1\right)$
$c=\{\begin{array}{cc}8& \text{if}b=6\\ 6& \text{if}b=8\end{array}$

Distance of the point $P(a,b,c)$ from $z$-axis $=\sqrt{a^2+b^2}$
$2\sqrt{41}=\sqrt{a^2+b^2}$
$\Rightarrow 164=a^2+b^2$
From eqn$\left(1\right),b$ can be either 8 or 6
$a=\{\begin{array}{cc}10& \text{if}b=8\\ \sqrt{128}& \text{if}b=6\end{array}$
$\because a\in N$
$\therefore a=10$
$\Rightarrow b=8,c=6$
So, the values of $(a,b,c)=(10,8,6)$

Hence, $a+b+c=24$