Is the tens digit of a three-digit positive integer p divisible by 3?

I approached this by using the rule that the sum of the digits must be equal to a multiple of 3 for a number to be a multiple of 3.

so let p=abc, where abc are digits.

The question then is: is b divisible by 3?

1)

In the cases of c=7, 8, or 9, b remains unchanged and we basically have

a+b+c-7 = a multiple of 3.

In cases where c=0 to 6, the digits become (a, b-1, 10+c-7) and we have

a+(b-1)+(10+c-7)=a+b+c+2 = a multiple of 3

This is insufficient to determine if b is 0, 3, 6, or 9 because

(a,b,c) can be, for example, (3,6,7) and (4,5,7) and satisfy the condition p-7 is a multiple of 3.

Insufficient.

2)
Analysis is similar for p-13=multiple of 3
In cases where c>=3, and b>=1, the digits become (a,b-1,c-3)
c>=3 and b=0, (a-1,9,c-3)
c<3 and b>=2, (a,b-2,10+c-3)
c<3 and b<=2, (a-1,10+b-2,10+c-3)

Notice we know nothing about a other than 1 is subtracted from it in the case of c>=3 and b=0. The problem is for all other cases, a can be adjusted up or down to make b a multiple of 3, or not.

Both together is still insufficient for the same reasons.