If the function $\mathrm{f}(x)=\left\{\begin{array}{cl}\frac{\cos a x-\cos b x}{\cos c x-\cos b x} & , \text { if } x \neq 0 \\ -1 & , \text { if } x=0\end{array}\right.$ is continuous at $x=0$, then $a^2, b^2, c^2$ are in

Continuity at $x = 0$ requires:

$\lim_{x \to 0} f(x) = f(0) = -1$

Evaluating the limit:

$\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos bx}$

Apply L'Hôpital's Rule twice:

$\lim_{x \to 0} \frac{-a \sin ax + b \sin bx}{-c \sin cx + b \sin bx} = \lim_{x \to 0} \frac{-a^2 \cos ax + b^2 \cos bx}{-c^2 \cos cx + b^2 \cos bx}$

Substituting $x = 0$:

$\frac{-a^2 \cdot 1 + b^2 \cdot 1}{-c^2 \cdot 1 + b^2 \cdot 1} = \frac{b^2 - a^2}{b^2 - c^2}$

Equate to $f(0) = -1$:

$\frac{b^2 - a^2}{b^2 - c^2} = -1$

$b^2 - a^2 = -(b^2 - c^2)$
$b^2 - a^2 = -b^2 + c^2$
$2b^2 = a^2 + c^2$

Thus $a^2$, $b^2$, and $c^2$ form an arithmetic progression with $b^2$ as the middle term.

\boxed{B}