Let $$a, b, c, d$$ and $$p$$ be non-zero distinct real numbers such that $$(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) = 0$$. Then:

We have the quadratic condition

$$\bigl(a^{2}+b^{2}+c^{2}\bigr)p^{2}-2\bigl(ab+bc+cd\bigr)p+\bigl(b^{2}+c^{2}+d^{2}\bigr)=0.$$

Because every term in the expression on the left is symmetric in the pairs $$(a,b)$$, $$(b,c)$$, $$(c,d)$$, it is natural to try to write it as a sum of three perfect squares. We write down those three candidates and expand them one by one.

First square:

$$(ap-b)^{2}=a^{2}p^{2}-2abp+b^{2}.$$

Second square:

$$(bp-c)^{2}=b^{2}p^{2}-2bcp+c^{2}.$$

Third square:

$$(cp-d)^{2}=c^{2}p^{2}-2cdp+d^{2}.$$

Now we add the three expansions term by term:

$$\begin{aligned} (ap-b)^{2}+(bp-c)^{2}+(cp-d)^{2} &=\bigl(a^{2}p^{2}+b^{2}p^{2}+c^{2}p^{2}\bigr) -2\bigl(abp+bcp+cdp\bigr) +\bigl(b^{2}+c^{2}+d^{2}\bigr)\\[4pt] &=\bigl(a^{2}+b^{2}+c^{2}\bigr)p^{2} -2\bigl(ab+bc+cd\bigr)p +\bigl(b^{2}+c^{2}+d^{2}\bigr). \end{aligned}$$

But the right-hand side of this last equality is exactly the quadratic expression given to be zero. Hence we have

$$ (ap-b)^{2}+(bp-c)^{2}+(cp-d)^{2}=0. $$

A sum of real squares can be zero only if each square itself is zero. So we must have the three simultaneous equations

$$\begin{cases} ap-b=0,\\[2pt] bp-c=0,\\[2pt] cp-d=0. \end{cases}$$

Solving them sequentially gives

$$b=ap,\qquad c=bp,\qquad d=cp.$$

Dividing each equality by the preceding term we obtain the equal ratios

$$\dfrac{b}{a}=p,\qquad \dfrac{c}{b}=p,\qquad \dfrac{d}{c}=p.$$

All three successive ratios are equal to the same non-zero real number $$p$$. Therefore

$$a,\;b,\;c,\;d$$

form a geometric progression with common ratio $$p$$. None of the other listed statements follows necessarily from the given condition.

Hence, the correct answer is Option C.