If the vectors $m \hat{i}+m \hat{j}+n \hat{k}, \hat{i}+\hat{k}, n \hat{i}+n \hat{j}+p \hat{k}$ lie in a plane then...

Given vectors $\mathrm{a} = m \hat{i} + m \hat{j} + n \hat{k}$, $\mathrm{b} = \hat{i} + \hat{k}$, and $\mathrm{c} = n \hat{i} + n \hat{j} + p \hat{k}$, coplanarity requires the scalar triple product to vanish.

The determinant condition yields

$\begin{vmatrix} m & m & n \ 1 & 0 & 1 \ n & n & p \end{vmatrix} = 0$

Expanding along the first row: $m(0 \cdot p - 1 \cdot n) - m(1 \cdot p - 1 \cdot n) + n(1 \cdot n - 0 \cdot n) = -mn - m(p - n) + n^2 = -mp + n^2 = 0$

This simplifies to $n^2 = mp$, indicating $m$, $n$, and $p$ form a geometric progression.

Among the options, only C satisfies this condition.