Scalar Triple Product in Determinant Form

$$M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}$$

$$\vec{u} = \alpha\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \gamma\hat{k}$$

$$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \beta\hat{j} + \delta\hat{k}$$

$$\vec{w} = -\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \mu\hat{k}$$

$$\alpha^2 + \frac{1}{3} + \gamma^2 = 1 \quad \therefore \alpha^2 + \gamma^2 = \frac{2}{3} \quad \dots(1)$$

$$\frac{1}{2} + \beta^2 + \delta^2 = 1 \quad \beta^2 + \delta^2 = \frac{1}{2} \quad \dots(2)$$

$$\frac{1}{2} + \frac{1}{3} + \mu^2 = 1 \quad \mu^2 = \frac{1}{6} \quad \dots(3)$$

$$\gamma^2 + \delta^2 + \mu^2 = 1$$

$$\gamma^2 + \delta^2 = 1 - \frac{1}{6} = \frac{5}{6}$$

For Q:

$$x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}$$

$$x\vec{u} \cdot \vec{u} + y\vec{v} \cdot \vec{u} + z\vec{w} \cdot \vec{u} = \vec{u} \cdot \hat{j}$$

$$x = \frac{1}{\sqrt{3}}$$

For R: 

$$\vec{u}, \vec{v}, \vec{w}$$ are mutually perpendicular unit vectors.

$$\vert{}\vec{u} \cdot (\vec{v} \times \vec{w})\vert{}$$

$$\begin{bmatrix} \vec{u} & \vec{v} & \vec{w} \end{bmatrix} = 1$$

For S: 

$$\vec{u} \times (\vec{v} \times \vec{w})$$

$$(\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$$

$$= 0 - 0 = 0$$

Step 1: Relate the scalar triple product of cross products to the base vectors
Let $$V_1$$ be the scalar triple product of the vectors $$\vec{p}$$, $$\vec{q}$$, and $$\vec{r}$$, which represents the volume of the parallelopiped formed by them:
$$V_1 = [\vec{p} \;\; \vec{q} \;\; \vec{r}]$$

The problem specifies the volume of the parallelopiped formed by their cross products, $$\vec{p} \times \vec{q}$$, $$\vec{q} \times \vec{r}$$, and $$\vec{r} \times \vec{p}$$. Using the standard scalar triple product identity for vector cross products:
$$[(\vec{p} \times \vec{q}) \;\; (\vec{q} \times \vec{r}) \;\; (\vec{r} \times \vec{p})] = [\vec{p} \;\; \vec{q} \;\; \vec{r}]^2 = V_1^2$$

We are given that this volume is equal to $$3136$$:
$$V_1^2 = 3136 \implies V_1 = \pm \sqrt{3136} = \pm 56$$

Step 2: Express the scalar triple product V_1 in terms of the base vectors
The vectors $$\vec{p}$$, $$\vec{q}$$, and $$\vec{r}$$ are expressed as linear combinations of the non-coplanar vectors $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$. We can write their scalar triple product using the determinant of their coefficients multiplied by the scalar triple product of the base vectors:
$$[\vec{p} \;\; \vec{q} \;\; \vec{r}] = \begin{vmatrix} 2 & 3 & -1 \\ 1 & -2 & 2 \\ 3 & 1 & \lambda \end{vmatrix} [\vec{a} \;\; \vec{b} \;\; \vec{c}]$$

We are given that $$[\vec{a} \;\; \vec{b} \;\; \vec{c}] = 4$$. Let's evaluate the coefficient determinant:
$$D = \begin{vmatrix} 2 & 3 & -1 \\ 1 & -2 & 2 \\ 3 & 1 & \lambda \end{vmatrix}$$
$$D = 2((-2)(\lambda) - (2)(1)) - 3((1)(\lambda) - (2)(3)) - 1((1)(1) - (-2)(3))$$
$$D = 2(-2\lambda - 2) - 3(\lambda - 6) - 1(1 + 6)$$
$$D = -4\lambda - 4 - 3\lambda + 18 - 7$$
$$D = -7\lambda + 7$$

Substituting $$D$$ and the base scalar triple product into the equation for $$V_1$$:
$$V_1 = (-7\lambda + 7) \times 4 = -28\lambda + 28$$

Step 3: Solve for the possible values of \lambda
From Step 1, we know that $$V_1 = \pm 56$$. This gives us two separate equations to solve:

Case A: When $$V_1 = 56$$
$$-28\lambda + 28 = 56$$
$$-28\lambda = 28 \implies \lambda_1 = -1$$

Case B: When $$V_1 = -56$$
$$-28\lambda + 28 = -56$$
$$-28\lambda = -84 \implies \lambda_2 = 3$$

Step 4: Calculate the sum of all possible values of \lambda
The problem asks for the product of all possible real values of $$\lambda$$:
$$\text{Product} = \lambda_1 \lambda_2 = -(1) (3) = -3$$

The volume of a parallelepiped equals the absolute value of the scalar triple product:
$$V = |[\vec{a}, \vec{b}, \vec{c}]| = |\vec{a} \cdot (\vec{b} \times \vec{c})|$$

This equals the absolute value of the determinant:
$$\begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix}$$

Expanding along the first row:
$$\det = 1 \begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} + (-1) \begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}$$

Calculating each $$2 \times 2$$ determinant:
$$\begin{vmatrix} -1 & 1 \\ 1 & 1 \end{vmatrix} = (-1)(1) - (1)(1) = -2$$

$$\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 1$$

$$\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (-1)(1) = 3$$

Computing the determinant:
$$\det = 1(-2) - 2(1) + (-1)(3) = -2 - 2 - 3 = -7$$

Therefore: $$V = |-7| = 7$$