The ages, in years, of four members of a family are represented
by a, b, c and d, where a < b < c < d. Their mean age is 34,
their median age is 33, and the range of their ages is 32.
What is the value of a?
The mean age is 34, so the sum of the four ages is $4 \times 34 = 136$. The median age is 33, which means the two middle ages must sum to $2 \times 33 = 66$.
The range is 32, so $d - a = 32$. Since $a < b < c < d$, we can also say that $d - c > 0$, $d - b > 0$, and $c - b > 0$. Therefore, \begin{align*}
d &> c > b > a. \\
d - c &> 0 \quad \Rightarrow \quad c < d. \\
d - b &> 0 \quad \Rightarrow \quad b < d. \\
c - b &> 0 \quad \Rightarrow \quad b < c.
\end{align*} Therefore, $d$ is the largest of the four ages, $c$ is the second largest, $b$ is the second smallest, and $a$ is the smallest.
Since $d$ is the largest, $d > a$, which means $d - a > 0$. Also, since $b$ is the second smallest, $b < c$.
We can already conclude that $c - a$ is at least $32 - 1 = 31$.
If we let $a = c - 31$, then $d = a + 32 = (c - 31) + 32 = c + 1$. Also, $b < c$ gives us $b < c - 31$, so $b - a < 0$. Therefore, $d - a = (c + 1) - (c - 31) = 32 > 0$. Hence, our values of $a$, $b$, $c$, and $d$ satisfy all the criteria.
The ages are $a = c - 31$, $b < c - 31$, $c$, and $d = c + 1$. The sum of these four ages is
\begin{align*}
(a + b + c + d) &= (c - 31) + b + c + (c + 1) \\
&= 3c - 30 + b + 1 \\
&= 3c + b - 19.
\end{align*}We are given that the sum of the ages is 136, so $$3c + b - 19 = 136,$$which means that $$3c + b = 155.$$Therefore, $3c + b$ is equal to 149, 152, 155, 158, 161, etc. In particular, $3c + b$ is a multiple of 3. However, the only value of $c$ for which $3c + b = 155$ is $c = 52$, which gives us a negative value of $b$. This means that there are no integers $a$, $b$, $c$, and $d$ that can satisfy the given conditions, so the answer is $\boxed{\text{none}}$.