Let $$X = \left[ {\matrix{ x \cr y \cr z \cr } } \right]$$

PX = O

$$\left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

x + 2y + z = 0........(1)

-2x + 3y – 4z = 0....(2)

x + 9y - z = 0..........(3)

from (1) & (3)
$$ \Rightarrow $$ 2x+11y =0

from (1) & (2)
$$ \Rightarrow $$ 2x + 11y = 0

from (2) & (3)
–6x –33y = 0
$$ \Rightarrow $$ 2x +11y = 0

putting value of x in (1), we get
–7y + 2z = 0

Now $${\left( {{{11y} \over 2}} \right)^2} + {y^2} + {\left( {{{7y} \over 2}} \right)^2} = 1$$

y²(121 + 1 + 49) = 4

y²(171) = 4

$$y = \pm {2 \over {\sqrt {171} }}$$

$$ \Rightarrow x = \pm {7 \over {\sqrt {171} }}$$

$$ \Rightarrow z = \pm {{11} \over {\sqrt {171} }}$$

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