Why in BFPRT (median of medians) algorithm the partition of the array by $7$ blocks would work but with the $3$ fail?
Let's take a look at the case of size 7 first. Here, we want to show a linear upper bound for $T(n)$.
Thus, we choose the recurrence relation $T(n) \leq T(n / 7) + T(5 / 7 n) + dn$ (remember, it is an upper bound).
We guess that the solution is of the form $T(n) \leq cn$ for some constant $c$ and prove it with induction:
$$T(n) \leq T(n / 7) + T(5 / 7 \cdot n) + dn \leq c \cdot n/7 + c \cdot 5/7 \cdot n + dn = 6/7 \cdot cn +dn. $$
To this end, we choose $c = 7 d$, resulting in $T(n) \leq cn$.

Now we want to show a lower bound for size 3 (for the worst-case, tho!).
So we have find a lower bound for its recurrence equation, namely:
$$T(n) \geq T(n/3) + T(2/3 \cdot n) + dn .$$
The list of medians takes $T(n/3)$ since there are $n/3$ medians (resp. blocks).
In the worst case, we have to recursively search in an area of size at least $2/3 n$.
Clearly, this recurrence relation is not linear.
To get an intuition for this, imagine that we would have equality. 
Assuming a solution of the form $T(n) = cn$, we get $T(n) = 1/3 \dot cn + 2/3 \dot cn + dn$.
Since $d&gt;0$, $T(n) &gt; cn$ which contradicts our assumption.

For more information, check out the analysis on Wikipedia.

The key part is that for block size $3$ the sizes of the recursive calls add up to $n$, which does not help to reduce the running time.

If you have more questions, feel free to ask!