Let $$\vec{w} = \hat{i} + \hat{j} - 2\hat{k}$$, and $$\vec{u}$$ and $$\vec{v}$$ be two vectors, such that $$\vec{u} \times \vec{v} = \vec{w}$$ and $$\vec{v} \times \vec{w} = \vec{u}$$. Let $$\alpha, \beta, \gamma$$, and $$t$$ be real numbers such that

$$\vec{u} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}, \quad -t\alpha + \beta + \gamma = 0, \quad \alpha - t\beta + \gamma = 0, \quad \text{and} \quad \alpha + \beta - t\gamma = 0.$$

Match each entry in List-I to the correct entry in List-II and choose the correct option.

The three vectors are connected by
$$\vec{u}\times\vec{v}= \vec{w},\qquad \vec{v}\times\vec{w}= \vec{u},\qquad \vec{w}= \hat{i}+ \hat{j}-2\hat{k}.$$ The cross-product of two vectors is perpendicular to each of them, so

$$\vec{u}\cdot\vec{v}=0, \qquad \vec{v}\cdot\vec{w}=0, \qquad \vec{w}\cdot\vec{u}=0.$$

Hence the three vectors are pair-wise perpendicular.

Magnitude of $$\vec{w}$$:
$$|\vec{w}|^{2}=1^{2}+1^{2}+(-2)^{2}=6.$$

Using the identity $$|\vec{a}\times\vec{b}|^{2}=|\,\vec{a}\,|^{2}|\,\vec{b}\,|^{2}-(\vec{a}\cdot\vec{b})^{2},$$
$$|\vec{w}|^{2}=|\vec{u}\times\vec{v}|^{2}=|\vec{u}|^{2}|\vec{v}|^{2}\quad(\because\vec{u}\cdot\vec{v}=0)$$ and
$$|\vec{u}|^{2}=|\vec{v}\times\vec{w}|^{2}=|\vec{v}|^{2}|\vec{w}|^{2}\quad(\because\vec{v}\cdot\vec{w}=0).$$

Let $$|\vec{u}|^{2}=a,\; |\vec{v}|^{2}=b.$$
Then $$6=ab \quad\text{and}\quad a=6b.$$ Substituting, $$6=(6b)b\;\Longrightarrow\; b^{2}=1\;\Longrightarrow\; b=1.$$ Therefore

Case P: $$|\vec{v}|^{2}=1\; \Rightarrow\; (P)\to(2).$$

The components of $$\vec{u}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$$ obey the three equations

$$-t\alpha+\beta+\gamma=0\; -(1)$$
$$\alpha-t\beta+\gamma=0\; -(2)$$
$$\alpha+\beta-t\gamma=0\; -(3).$$

For a non-trivial solution, the determinant of the coefficient matrix must vanish:

$$\begin{vmatrix}-t&1&1\\[2pt]1&-t&1\\[2pt]1&1&-t\end{vmatrix}=0 \;\Longrightarrow\; -t^{3}+3t+2=0 \;\Longrightarrow\; t^{3}-3t-2=0 =(t-2)(t+1)^{2}=0.$$

Thus $$t=2\quad\text{or}\quad t=-1.$$ We must also satisfy $$\vec{w}\cdot\vec{u}=0\;\Rightarrow\; \alpha+\beta-2\gamma=0\; -(4).$$

Taking $$t=-1$$
Equations (1)-(3) all reduce to $$\alpha+\beta+\gamma=0.$$ With $$\alpha=\sqrt{3}$$, $$\beta+\gamma=-\sqrt{3}\; -(5).$$ Using (4): $$\sqrt{3}+\beta-2\gamma=0 \;\Longrightarrow\; -3\gamma=0 \;\Longrightarrow\; \gamma=0,\;\beta=-\sqrt{3}.$$ Hence

Case Q: $$\gamma^{2}=0\;\Rightarrow\; (Q)\to(1).$$

Case R: $$(\beta+\gamma)^{2}=(-\sqrt{3})^{2}=3\;\Rightarrow\; (R)\to(4).$$

Taking $$t=2$$
From (1): $$\beta=2\alpha-\gamma.$$ From (2): $$\alpha-2\beta+\gamma=0 \;\Longrightarrow\; \gamma=\alpha.$$ Therefore $$\beta=\alpha.$$ With $$\alpha=\sqrt{2}$$ we get $$t=2,$$ hence

Case S: $$t+3=2+3=5\;\Rightarrow\; (S)\to(5).$$

Collecting all matches:
(P)$$\to$$(2), (Q)$$\to$$(1), (R)$$\to$$(4), (S)$$\to$$(5).

Thus the correct option is
Option A which is: (P)$$\to$$(2), (Q)$$\to$$(1), (R)$$\to$$(4), (S)$$\to$$(5).