The ends $AB$ of a straight line segment of constant length $c$ slide upon the fixed rectangular axes $OX$ and $OY$ respectively. If the rectangle $OAPB$ be completed, then the locus of the foot of perpendicular drawn from $P$ to $AB$ is 

The diagram for the given details is drawn.

Let $A\equiv (a,0),B\equiv (0,b)$ then, $P\equiv (a,b)$.

Since,

$AB=c$

$$\begin{gathered} \sqrt{a^2+b^2}=c \\ a^2+b^2=c^2 \end{gathered}$$

$\text{Let }Q\equiv \left(x_1,y_1\right)$ and $PQ\perp AB$.

 Slope of $PQ\times$ Slope of $AB=-1$

$\left(\frac{b-y_1}{a-x_1}\right)\times \left(\frac{b-0}{0-a}\right)=-1$

$a{x}_1-b{y}_1=a^2-b^2 .......\left(1\right)$

Equation of $AB$ is $\frac{x}{a}+\frac{y}{b}=1$

But $Q$ lies on $AB$ then $\frac{x_1}{a}+\frac{y_1}{b}=1$

$b{x}_1+a{y}_1=ab ......\left(2\right)$

From equation $\left(1\right) and \left(2\right)$,

$x_1=\frac{a^3}{a^2+b^2},y_1=\frac{b^3}{a^2+b^2}$

Now, $x_1^{2/3}+y_1^{2/3}=\frac{\left(a^2+b^2\right)}{{\left(a^2+b^2\right)}^{2/3}}$

                            $$\begin{gathered} ={\left(a^2+b^2\right)}^{1/3} \\ =c^{2/3} \end{gathered}$$

Hence, the required locus is $x^{2/3}+y^{2/3}=c^{2/3}$.

Therefore, the correct answer is option $\left(3\right)$.