Statement 1: If the points $$(1, 2, 2)$$, $$(2, 1, 2)$$ and $$(2, 2, z)$$ and $$(1, 1, 1)$$ are coplanar, then $$z = 2$$. Statement 2: If the 4 points $$P, Q, R$$ and $$S$$ are coplanar, then the volume of the tetrahedron $$PQRS$$ is $$0$$.

For four points $$A(x_1,y_1,z_1),\;B(x_2,y_2,z_2),\;C(x_3,y_3,z_3)$$ and $$D(x_4,y_4,z_4)$$ to be coplanar, the scalar triple product of the vectors $$\overrightarrow{AB},\;\overrightarrow{AC},\;\overrightarrow{AD}$$ must be zero:

$$\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD})=0$$

This follows because the volume of tetrahedron $$ABCD$$ is
$$V=\frac16\left|\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD})\right|,$$
and volume is zero precisely when all four points lie in one plane.

Let
$$A(1,2,2),\;B(2,1,2),\;C(2,2,z),\;D(1,1,1).$$

Compute the three vectors based at $$A$$:

$$\overrightarrow{AB}=B-A=(2-1,\,1-2,\,2-2)=(1,\,-1,\,0),$$
$$\overrightarrow{AC}=C-A=(2-1,\,2-2,\,z-2)=(1,\,0,\,z-2),$$
$$\overrightarrow{AD}=D-A=(1-1,\,1-2,\,1-2)=(0,\,-1,\,-1).$$

Write the scalar triple product as the determinant

$$ \overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD}) =\begin{vmatrix} 1 & -1 & 0 \\ 1 & 0 & z-2 \\ 0 & -1 & -1 \end{vmatrix}. $$

Expanding along the first row:

$$ \begin{aligned} \det &= 1\Bigl| \begin{matrix} 0 & z-2\\ -1 & -1 \end{matrix} \Bigr| -(-1)\Bigl| \begin{matrix} 1 & z-2\\ 0 & -1 \end{matrix} \Bigr| +0\cdot(\text{minor})\\[4pt] &=1\left(0\cdot(-1)-(z-2)(-1)\right)-(-1)\left(1\cdot(-1)-(z-2)\cdot0\right)\\[4pt] &=(z-2)-(-1)(-1)\\[4pt] &=(z-2)-1\\[4pt] &=z-3. \end{aligned} $$

The coplanarity condition $$\det=0$$ therefore gives $$z-3=0 \implies z=3.$$

Statement 1 claims $$z=2$$; this contradicts the derived value $$z=3$$. Hence Statement 1 is false.

Statement 2 asserts: “If the four points $$P,Q,R,S$$ are coplanar, the volume of tetrahedron $$PQRS$$ is zero.”
Since volume equals one-sixth the magnitude of the same scalar triple product used above, coplanarity indeed forces the volume to vanish. Statement 2 is therefore true.

Thus,
Statement 1: false;
Statement 2: true.

Option A which is: Statement 1 is false, Statement 2 is true.