Observation: The vectors $\vec{u},\vec{v},\vec{w}$ are exactly the columns of $M$. Since $M{M}^T=I$, $M$ is orthogonal, so $M^{T}M=I$ as well, meaning the columns are mutually orthogonal unit vectors.

Row-norm equations from $M{M}^T=I$:

Row 1: ${\alpha }^2+1/2+1/2=1\Rightarrow {\alpha }^2=0\Rightarrow \alpha =0$.

Wait, this contradicts column 1 being a unit vector. Re-checking: actually $M{M}^T=I$ says rows are orthonormal. Row 1 norm: ${\alpha }^2+1/2+1/2=1\Rightarrow \alpha =0$. But the columns also form an orthonormal set (since $M$ orthogonal), and column 1 has norm ${\alpha }^2+1/3+{\gamma }^2=1$.

Using column-orthonormality (since $M$ is orthogonal):

$|\vec{u}{|}^2={\alpha }^2+1/3+{\gamma }^2=1$, $|\vec{v}{|}^2=1/2+{\beta }^2+{\delta }^2=1$, $|\vec{w}{|}^2=1/2+1/3+{\mu }^2=1\Rightarrow {\mu }^2=1/6$.

From row-orthonormality, the column norms come out the same way: ${\alpha }^2+{\gamma }^2=2/3$, ${\beta }^2+{\delta }^2=1/2$.

Combined with row 3 norm ${\gamma }^2+{\delta }^2+{\mu }^2=1$: ${\gamma }^2+{\delta }^2=1-1/6=5/6$. P $\to$ 5.

(Q) Since $\vec{u},\vec{v},\vec{w}$ are mutually perpendicular unit vectors, dotting $x\vec{u}+y\vec{v}+z\vec{w}=\hat{j}$ with $\vec{u}$ gives $x=\vec{u}\cdot \hat{j}=\frac{1}{\sqrt{3}}$. Q $\to$ 4.

(R) $|\vec{u}\cdot (\vec{v}\times \vec{w})|$ is the absolute value of the scalar triple product, which equals the absolute determinant of $M$. Since $M$ is orthogonal, $|\det M|=1$. R $\to$ 2.

(S) Using the vector triple product identity, $\vec{u}\times (\vec{v}\times \vec{w})=(\vec{u}\cdot \vec{w})\vec{v}-(\vec{u}\cdot \vec{v})\vec{w}$. Both dot products vanish by mutual perpendicularity, so the magnitude is $0$. S $\to$ 1.

$\therefore$ The correct option is (A).