3. In the Cartesian coordinate system $x O y$, it is known that there are three points $A(a, 1), B(2, b), C(3,4)$.
If the projections of $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are the same, then $3 a-4 b=$ $\qquad$
3. 2 .

Solution 1 The projections of vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are $\frac{\overrightarrow{O A} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}, \frac{\overrightarrow{O B} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}$, respectively.
According to the problem, we have
$$
\overrightarrow{O A} \cdot \overrightarrow{O C}=\overrightarrow{O B} \cdot \overrightarrow{O C} \text {, }
$$

which means $3 a+4=6+4 b$.
Thus, $3 a-4 b=2$.
Solution 2 Since the projections of vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are the same, $A B \perp O C$, that is,
$$
\overrightarrow{A B} \cdot \overrightarrow{O C}=0 \text {. }
$$

Therefore, $3(2-a)+4(b-1)=0$, which simplifies to $3 a-4 b=2$.