Determine the number of positive integral values of $$p$$ for which there exists a triangle with sides $$a$$, $$b$$, and $$c$$ which satisfy $$a^{2}+(p^{2}+9)b^{2}+9c^{2}-6ab-6pbc=0$$.

The given condition is
$$a^{2}+(p^{2}+9)b^{2}+9c^{2}-6ab-6pbc=0 \quad\quad -(1)$$
where $$a,\,b,\,c$$ are the three side-lengths of a triangle and $$p$$ is a positive integer.

Step 1 : Convert the left side of (1) into a sum of perfect squares

Group the terms involving $$a$$ and $$b$$ first:

$$a^{2}-6ab+9b^{2}=(a-3b)^{2}.$$

The remainder in (1) is $$p^{2}b^{2}+9c^{2}-6pbc$$ which factors similarly:

$$p^{2}b^{2}+9c^{2}-6pbc=(pb-3c)^{2}.$$

Hence (1) becomes

$$(a-3b)^{2}+(pb-3c)^{2}=0 \quad\quad -(2)$$

Step 2 : Deduce the equalities implied by (2)

Both squares in (2) are non-negative; their sum is zero only when each square is zero. Therefore

$$a-3b=0 \;\Rightarrow\; a=3b,$$

$$pb-3c=0 \;\Rightarrow\; c=\dfrac{p}{3}\,b.$$

Thus the three sides are proportional to

$$a:b:c = 3b:b:\dfrac{p}{3}b \; \Longrightarrow \; 3:1:\dfrac{p}{3}.$$

Step 3 : Impose the triangle inequalities

Let the three side lengths be $$3,\;1,\;\dfrac{p}{3}$$ (any common positive scale cancels).
For a triangle we need the sum of any two sides to exceed the third:

1. $$3+1 \gt \dfrac{p}{3} \;\Longrightarrow\; 4 \gt \dfrac{p}{3} \;\Longrightarrow\; p \lt 12.$$

2. $$1+\dfrac{p}{3} \gt 3 \;\Longrightarrow\; \dfrac{p}{3} \gt 2 \;\Longrightarrow\; p \gt 6.$$

3. $$3+\dfrac{p}{3} \gt 1$$ is automatically true for every positive $$p$$.

Combining (1) and (2):

$$6 \lt p \lt 12.$$

Step 4 : Count the admissible positive integers

The integers that satisfy $$6 \lt p \lt 12$$ are
$$p = 7,\,8,\,9,\,10,\,11.$$

There are exactly five such positive integral values of $$p$$.

Answer: 05