$\because \left|P\right|=0$

$\therefore$ System of equations 

$x+2 y+z=0$ 

$2x-3y+4z=0$ 

$x+9y-z=0$ has infinitely many solutions.

Let $z=k\in R$ then $x=-\frac{11\mathrm{k}}{7}, \mathrm{y}=\frac{2\mathrm{k}}{7}, \mathrm{z}=\mathrm{k}$ 

Also $x^2+y^2+z^2=1$ 

$\therefore \mathrm{k}=\pm \frac{7}{\sqrt{174}}$ 

$\therefore$ Two solutions are only possible.

Hence, set $A$ contains exactly two elements.