# Let $A = \left \{ X = (x,y,z)^{2} : Px = 0 , and , x^{2}+y^{2}+z^{2}=1 \right \}$ where $P = \begin{pmatrix} 1 & 2 &1 \\ -2& 3 &-4 \\ 1& 9 &-1 \end{pmatrix}$ then Set A  contains  : Option: 1 exactly two elements   Option: 2 is an empty set , Option: 3  is a singletan set ,   Option: 4 contains more than 2 elements .

$\\\text { Given } P=\left[\begin{array}{ccc} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{array}\right],\\ |\mathrm{P}|=1(3\times (-1)-(-4)\times 9)-2((-1)\times(-2)-(-4)\times1)+1(9\times(-2)-3\times 1)\\|P|=0 \\\text {Here }|P|=0\;\; \\ \text {also given } P X=0$

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

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

By solving these equation

\begin{aligned}{l} \text { we get } x=\frac{-11 \lambda}{2} ; y=\lambda ; z=\frac{7 \lambda}{2} \\ \text { 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}}} \end{aligned}

So, there are 2 values of $\lambda$

Therefore,  there are 2 solution set of (x, y, z).

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