Let $$A = \left\{X = (x, y, z)^T : PX = 0 \text{ and } x^2 + y^2 + z^2 = 1\right\}$$ where $$P = \begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix}$$ then the set $$A$$:

We have to solve the simultaneous conditions

$$PX = 0 \qquad\text{and}\qquad x^{2}+y^{2}+z^{2}=1,$$

where

$$P=\begin{bmatrix}1&2&1\\[2pt]-2&3&-4\\[2pt]1&9&-1\end{bmatrix},\qquad X=\begin{bmatrix}x\\y\\z\end{bmatrix}.$$

First we impose the homogeneous system $$PX=0.$$ Writing this matrix equation in component form gives three linear equations:

$$\begin{aligned} 1)\;&x+2y+z&=0,\\ 2)\;-2x+3y-4z&=0,\\ 3)\;x+9y-z&=0. \end{aligned}$$

From the first equation we can express $$x$$ in terms of $$y$$ and $$z$$:

$$x=-2y-z. \quad -(4)$$

We now substitute (4) into the second equation. We get

$$-2(-2y-z)+3y-4z=0.$$

Expanding gives

$$4y+2z+3y-4z=0,$$

which simplifies to

$$7y-2z=0.$$

Hence

$$z=\frac{7}{2}y. \quad -(5)$$

Next we substitute (4) and (5) into the third equation:

$$x+9y-z=0$$

becomes

$$\Bigl(-2y-\frac{7}{2}y\Bigr)+9y-\frac{7}{2}y=0.$$

The left‐hand side equals

$$\left(-2-\frac{7}{2}\right)y+9y-\frac{7}{2}y =\left(-\frac{11}{2}\right)y+9y-\frac{7}{2}y =-\frac{11}{2}y-\frac{7}{2}y+9y =-\frac{18}{2}y+9y =-9y+9y=0,$$

so the third equation is automatically satisfied. Therefore the solution set of $$PX=0$$ is one‐dimensional. We may take $$y=t$$ as a free parameter. Using (5) we have $$z=\dfrac{7}{2}t,$$ and then (4) gives $$x=-2t-\dfrac{7}{2}t=-\dfrac{11}{2}t.$$

Thus every solution vector can be written as

$$X=t\begin{bmatrix}-\dfrac{11}{2}\\[4pt]1\\[4pt]\dfrac{7}{2}\end{bmatrix} =\left(\frac{t}{2}\right)\begin{bmatrix}-11\\2\\7\end{bmatrix}.$$

Let us denote

$$\begin{bmatrix}-11\\2\\7\end{bmatrix}=v.$$

Then every vector in the null space is $$X=s\,v,$$ where $$s=\dfrac{t}{2}\in\mathbb R.$$

Now we impose the unit‐length condition $$x^{2}+y^{2}+z^{2}=1.$$ For $$X=s\,v$$ this becomes

$$s^{2}\bigl((-11)^{2}+2^{2}+7^{2}\bigr)=1.$$

We calculate the squared length of $$v$$:

$$(-11)^{2}+2^{2}+7^{2}=121+4+49=174.$$

Hence

$$s^{2}\cdot174=1 \quad\Longrightarrow\quad s^{2}=\frac{1}{174}.$$

Taking square roots,

$$s=\frac{1}{\sqrt{174}}\quad\text{or}\quad s=-\frac{1}{\sqrt{174}}.$$

Therefore there are exactly two vectors satisfying both conditions, namely

$$X_{1}= \frac{1}{\sqrt{174}}\begin{bmatrix}-11\\2\\7\end{bmatrix},\qquad X_{2}=-\frac{1}{\sqrt{174}}\begin{bmatrix}-11\\2\\7\end{bmatrix}.$$

Thus the set $$A$$ contains precisely two elements.

Hence, the correct answer is Option D.