Given $P=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & 1\end{array}\right],$ Here $|\mathrm{P}|=0 $ and also given $P X=0$

$\Rightarrow\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=0$

$\left.\begin{array}{l}x+2 y+z=0 \\ \Rightarrow-2 x+3 y-4 z=0 \\ x+9 y-z=0\end{array}\right\} \quad-\quad D=0,$ so system have infinite many solutions,

By solving these equation

we get $x=\frac{-11 \lambda}{2} ; y=\lambda ; z=\frac{7 \lambda}{2}$

Also given, $x^{2}+y^{2}+z^{2}=1$

$\Rightarrow\left(\frac{-11 \lambda}{2}\right)^{2}+(\lambda)^{2}+\left(\frac{7 \lambda}{2}\right)^{2}=1$

$\Rightarrow \lambda=\pm \frac{1}{\sqrt{\frac{121}{4}+1+\frac{49}{4}}}$

so, there are 2 values of $\lambda$.

$\therefore$ so, there are 2 solution set of $(x, y, z) .$