For real numbers α, β, γ, and μ , consider the matrix M = bmatrix α...

Since $MM^T = I$, $M$ is an orthogonal matrix.

The columns of an orthogonal matrix form an orthonormal basis for $\mathbb{R}^3$. The given vectors $\vec{u}, \vec{v}, \vec{w}$ are exactly the columns of $M$. Therefore, $\vec{u}, \vec{v}, \vec{w}$ are mutually orthogonal unit vectors.

For (P): The rows of $M$ are also orthonormal. The third row is $\begin{bmatrix} \gamma & \delta & \mu \end{bmatrix}$, so $\gamma^2 + \delta^2 + \mu^2 = 1$.

The third column $\vec{w}$ is a unit vector, so $|\vec{w}|^2 = \left(-\dfrac{1}{\sqrt{2}}\right)^2 + \left(\dfrac{1}{\sqrt{3}}\right)^2 + \mu^2 = 1$.

$\dfrac{1}{2} + \dfrac{1}{3} + \mu^2 = 1 \Rightarrow \mu^2 = \dfrac{1}{6}$

Substituting $\mu^2$ into the row equation gives $\gamma^2 + \delta^2 = 1 - \dfrac{1}{6} = \dfrac{5}{6}$. Thus, (P) $\rightarrow$ (5).

For (Q): Given $x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}$. Taking the dot product with $\vec{u}$ on both sides gives $x(\vec{u} \cdot \vec{u}) + y(\vec{v} \cdot \vec{u}) + z(\vec{w} \cdot \vec{u}) = \hat{j} \cdot \vec{u}$.

Since the vectors are orthonormal, $\vec{u} \cdot \vec{u} = 1$ and $\vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u} = 0$.

Thus, $x = \hat{j} \cdot \vec{u}$. From the definition of $\vec{u}$, the $\hat{j}$ component is $\dfrac{1}{\sqrt{3}}$, so $x = \dfrac{1}{\sqrt{3}}$. Thus, (Q) $\rightarrow$ (4).

For (R): The scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w})$ represents the determinant of the matrix formed by these column vectors, which is $M$.

Since $M$ is orthogonal, $\det(M) = \pm 1$. Thus, $|\vec{u} \cdot (\vec{v} \times \vec{w})| = |\det(M)| = 1$. Thus, (R) $\rightarrow$ (2).

For (S): Using the vector triple product expansion, $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$.

Since $\vec{u}, \vec{v}, \vec{w}$ are mutually orthogonal, $\vec{u} \cdot \vec{w} = 0$ and $\vec{u} \cdot \vec{v} = 0$.

Thus, $\vec{u} \times (\vec{v} \times \vec{w}) = \vec{0}$, and its magnitude is $0$. Thus, (S) $\rightarrow$ (1).

Answer: (P) $\rightarrow$ (5), (Q) $\rightarrow$ (4), (R) $\rightarrow$ (2), (S) $\rightarrow$ (1)