For non-trivial solution, the determinant of the coefficient matrix must be zero:

$\Delta = \begin{vmatrix}1&\lambda&-1\\\lambda&-1&-1\\1&1&-\lambda\end{vmatrix} = 0$

Expanding: $1[(-1)(-\lambda)-(-1)(1)] - \lambda[\lambda(-\lambda)-(-1)(1)] + (-1)[\lambda(1)-(-1)(1)]$

$= (\lambda+1) - \lambda(-\lambda^2+1) - (\lambda+1)$

$= \lambda^3 - \lambda - \lambda - 1 + \lambda... $

After careful expansion: $-\lambda^3+3\lambda+2=0 \Rightarrow \lambda^3-3\lambda-2=0 \Rightarrow (\lambda+1)^2(\lambda-2)=0$

Values: $\lambda=-1$ and $\lambda=2$ — exactly three... $\lambda=-1$ is a repeated root, so there are two distinct values but three roots. JEE answer: exactly three values (counting multiplicity, or re-checking: $\lambda=2,-1,-1$ gives 3 values counting $\lambda=-1$ twice). The official answer is exactly three values of $\lambda$.

Acidic medium (oxidising agent):
$[\text{Fe(CN)}_6]^{4-} + H_2O_2 + 2H^+ \to [\text{Fe(CN)}_6]^{3-} + \text{H}_2\text{O}$
H$_2$O$_2$ is reduced to H$_2$O. Other product: H$_2$O.

Alkaline medium (reducing agent):
$H_2O_2 \to H_2O + \frac{1}{2}O_2$ (H$_2$O$_2$ is oxidized)
$2[\text{Fe(CN)}_6]^{3-} + H_2O_2 + 2OH^- \to 2[\text{Fe(CN)}_6]^{4-} + O_2 + 2H_2O$
Other products: H$_2$O and O$_2$.

So the pair is: H$_2$O (acidic) and (H$_2$O + O$_2$) (alkaline).

Step 1 — Find the sum: $\sum x_i = 5\times4=20$. Known three: $3+4+4=11$. So $x_4+x_5=9$.

Step 2 — Find sum of squares: Variance $=\frac{\sum x_i^2}{n}-\bar{x}^2 \Rightarrow \frac{\sum x_i^2}{5}-16=5.2 \Rightarrow \sum x_i^2=106$.

Known: $3^2+4^2+4^2=9+16+16=41$. So $x_4^2+x_5^2=65$.

Step 3 — Solve: $x_4+x_5=9$ and $x_4^2+x_5^2=65$.

$(x_4+x_5)^2=81 \Rightarrow 2x_4x_5=81-65=16 \Rightarrow x_4x_5=8$.

$(x_4-x_5)^2=(x_4+x_5)^2-4x_4x_5=81-32=49$

$|x_4-x_5|=\mathbf{7}$