Scalar Triple Product

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 $$\lambda$$:
$$\text{Product} = \lambda_1 \lambda_2 = -(1) (3) = -3$$

Let

$$V=[\vec a\ \vec b\ \vec c]$$

denote the scalar triple product.

Given

$$|V|=5$$

Now

$$\vec p=\vec a\times\vec b,\qquad \vec q=\vec b\times\vec c,\qquad \vec r=\vec c\times\vec a$$

Using the identity

$$[\vec a\times\vec b,\ \vec b\times\vec c,\ \vec c\times\vec a]=[\vec a\ \vec b\ \vec c]^2$$

we get

$$[\vec p\ \vec q\ \vec r]=V^2$$

Hence

$$|[\vec p\ \vec q\ \vec r]|=25$$

Also,

$$\vec x=\vec p-2\vec q,\qquad \vec y=\vec q-2\vec r,\qquad \vec z=\vec r-2\vec p$$

Therefore

$$[\vec x\ \vec y\ \vec z]=\begin{vmatrix}1&0&-2\\-2&1&0\\0&-2&1\end{vmatrix}[\vec p\ \vec q\ \vec r]$$

Let

$$M=\begin{bmatrix}1&0&-2\\-2&1&0\\0&-2&1\end{bmatrix}$$

Then

$$[\vec x\ \vec y\ \vec z]=\det(M)\,[\vec p\ \vec q\ \vec r]$$

Now,

$$\det(M)=1\begin{vmatrix}1&0\\-2&1\end{vmatrix}+(-2)\begin{vmatrix}-2&1\\0&-2\end{vmatrix}$$

$$=1(1)+(-2)(4)$$

$$=-7$$

Hence

$$|[\vec x\ \vec y\ \vec z]|=|-7|\cdot25$$

$$=175$$

Therefore the required volume is

$$175$$

Final Answer :

$$175$$