Let w = î + ĵ − 2k̂, and u and v be two vectors such that u × v = w and v × w = u. Let α, β, γ, and t be real numbers such that

u = αî + βĵ + γk̂, −tα + β + γ = 0, α − tβ + γ = 0, and α + β − tγ = 0.

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

List-I	List-II
(P) |v|² is equal to	(1) 0
(Q) If α = √3, then γ² is equal to	(2) 1
(R) If α = √3, then (β + γ)² is equal to	(3) 2
(S) If α = √2, then t + 3 is equal to	(4) 3
	(5) 5

The correct option is (P) → (2), (Q) → (1), (R) → (4), (S) → (5).

Given:
$\mathbf{w}=\mathbf{i}+\mathbf{j}-2\mathbf{k}$
And two vectors $\mathbf{u}$ and $\mathbf{v}$ satisfy:
1) $\mathbf{u}\times \mathbf{v}=\mathbf{w}$
2) $\mathbf{v}\times \mathbf{w}=\mathbf{u}$

From equation (1), $\mathbf{w}$ is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$. Therefore, the angle between $\mathbf{u}$ and $\mathbf{v}$ is 90°.
Taking the magnitude of equation (1):
$\left|\mathbf{w}\right|=\left|\mathbf{u}\right|\left|\mathbf{v}\right|sin(90°)=\left|\mathbf{u}\right|\left|\mathbf{v}\right|$
Similarly, from equation (2), since $\mathbf{v}$ is perpendicular to $\mathbf{w}$, taking the magnitude gives:
$\left|\mathbf{u}\right|=\left|\mathbf{v}\right|\left|\mathbf{w}\right|$
Substituting $\left|\mathbf{u}\right|$ into the first magnitude equation:
$\left|\mathbf{w}\right|=\left(\right|\mathbf{v}\left|\right|\mathbf{w}\left)\right|\mathbf{v}|=|\mathbf{v}{|}^2\left|\mathbf{w}\right|$
Since $\mathbf{w}\ne 0$, we have:
$|\mathbf{v}{|}^2=1$
This matches (P) → (2).

Next, the magnitude of $\mathbf{w}$ is:
$\left|\mathbf{w}\right|=\sqrt{1^2+1^2+(-2{)}^2}=\sqrt{6}$
Thus, the magnitude of $\mathbf{u}$ is:
$\left|\mathbf{u}\right|=\left|\mathbf{v}\right|\left|\mathbf{w}\right|=1·\sqrt{6}=\sqrt{6}$
Since $\mathbf{u}=\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k}$, we have:
${\alpha }^2+{\beta }^2+{\gamma }^2=6$

We are given a system of homogeneous linear equations in $\alpha ,\beta ,\gamma$:
$-t\alpha +\beta +\gamma =0$
$\alpha -t\beta +\gamma =0$
$\alpha +\beta -t\gamma =0$
For a non-trivial solution (since $\left|\mathbf{u}\right|=\sqrt{6}\ne 0$), the determinant of the coefficient matrix must be zero:
$\left|\begin{array}{ccc}-t& 1& 1\\ 1& -t& 1\\ 1& 1& -t\end{array}\right|=0$
Expanding this determinant:
$-t(t^2-1)-1(-t-1)+1(1+t)=0$
$-t^3+3t+2=0$
$t^3-3t-2=0$
Factoring this cubic equation gives:
$(t+1{)}^2(t-2)=0$
Thus, the possible values of $t$ are $t=-1$ or $t=2$.

Case 1: If $t=2$
Substituting $t=2$ into the system of equations gives:
$\alpha =\beta =\gamma$
Since ${\alpha }^2+{\beta }^2+{\gamma }^2=6$, we have:
$3{\alpha }^2=6\Rightarrow \alpha =\pm \sqrt{2}$
Thus, if $\alpha =\sqrt{2}$, then $t=2$.
Therefore, $t+3=2+3=5$.
This matches (S) → (5).

Case 2: If $t=-1$
Substituting $t=-1$ into the system equations, they all simplify to:
$\alpha +\beta +\gamma =0$
If $\alpha =\sqrt{3}$, then:
$\beta +\gamma =-\sqrt{3}$
Squaring both sides:
$(\beta +\gamma {)}^2=3$
This matches (R) → (4).

Additionally, since $\mathbf{u}$ and $\mathbf{w}$ are perpendicular, their dot product is zero:
$\mathbf{u}·\mathbf{w}=0\Rightarrow \alpha +\beta -2\gamma =0$
Subtracting this from $\alpha +\beta +\gamma =0$:
$3\gamma =0\Rightarrow \gamma =0$
Thus, ${\gamma }^2=0$.
This matches (Q) → (1).