$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ which is unit distance from the origin cuts the coordinates axes at $A, B$ and $C$.
$\therefore \quad 1=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}} \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1$
Coordinate of vertices of $\triangle A B C$ is $A(a, 0,0), B(0, b, 0), C(0,0, c)$
$\because$ Centroid of $\triangle A B C$ is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$
Now, $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$ satisfied the equation
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$
$\therefore \quad \frac{9}{a^2}+\frac{9}{b^2}+\frac{9}{c^2}=k$
$9\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=k$ $\left[\because \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1\right]$
$\therefore \quad k=9$