An auditorium has seats numbered sequentially from 1 to 300 . Seats...

Total seats are numbered $1$ to $300$.

The VIP seats are those whose numbers are multiples of $3$.
Let $V$ be the set of VIP seat numbers:
$V = \{3, 6, 9, \dots, 300\}$

A VIP seat is defective if its number $n$ satisfies $\text{gcd}(n, 35) > 1$. Since $35 = 5 \times 7$, a seat is defective if its number is divisible by $5$ or $7$.
We need to find the sum of valid VIP seats, which are multiples of $3$ but not divisible by $5$ or $7$.

Using the Principle of Inclusion-Exclusion, the required sum is:
$S_{\text{valid}} = S_3 - S_{15} - S_{21} + S_{105}$

Sum of multiples of $3$ up to $300$:
$S_3 = 3 + 6 + \dots + 300 = 3 \times \dfrac{100 \times 101}{2} = 15150$

Sum of multiples of $15$ (i.e., $3 \times 5$) up to $300$:
$S_{15} = 15 + 30 + \dots + 300 = 15 \times \dfrac{20 \times 21}{2} = 3150$

Sum of multiples of $21$ (i.e., $3 \times 7$) up to $300$:
The largest multiple of $21$ up to $300$ is $294 = 21 \times 14$.
$S_{21} = 21 + 42 + \dots + 294 = 21 \times \dfrac{14 \times 15}{2} = 2205$

Sum of multiples of $105$ (i.e., $3 \times 5 \times 7$) up to $300$:
The multiples are $105$ and $210$.
$S_{105} = 105 + 210 = 315$

Therefore, the sum of all valid VIP seats is:
$S_{\text{valid}} = 15150 - 3150 - 2205 + 315 = 10110$

Answer: $10110$