Let $\overrightarrow{b}=4\widehat{i}+3\widehat{j}$ and $\overrightarrow{c}$ be be two vectors perpendicular to each other in the xy-plane. The dot product of
.the vector in the same plane having projection 1 and 2 along $\overrightarrow{b}$ and $\overrightarrow{c}$ with vector $3\widehat{i}+\widehat{j}$ is

Let  $\overrightarrow{c}=\alpha \widehat{i}+\beta \widehat{j}$

Given that $\overrightarrow{b}\perp \overrightarrow{c}\therefore \overrightarrow{b}\cdot \overrightarrow{c}=0$

$\begin{array}{l}\Rightarrow (4\widehat{i}+3\widehat{j})\cdot (\alpha \widehat{i}+\beta \widehat{j})=0\\ \Rightarrow 4\alpha +3\beta =0\Rightarrow \frac{\alpha }{3}=\frac{\beta }{-4}=\lambda \\ \Rightarrow \alpha =3\lambda ,\beta =-4\lambda ...\left(i\right)\end{array}$

Now let $\overrightarrow{a}=x\widehat{i}+y\widehat{j}$ be the required vectors.

Projection of  $\overrightarrow{a}$ along $\overrightarrow{b}$ gives $\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\frac{4x+3y}{\sqrt{4^2+3^2}}=1$

or $4x+3y=5 \dots \left(ii\right)$            

Also projection of  $\overrightarrow{a}$ along $\overrightarrow{b}$ gives

$\begin{array}{l}\frac{\overrightarrow{a}\cdot \overrightarrow{c}}{\left|\overrightarrow{c}\right|}=2\Rightarrow \frac{\alpha x+\beta y}{\sqrt{{\alpha }^2+{\beta }^2}}=2\\ \Rightarrow 3\lambda x-4\lambda y=10\lambda \Rightarrow 3x-4y=10 ...\left(iii\right)\end{array}$

Solving (ii) and (iii), we get $x=2, y=-1$

Therefore, the required vector is $2\widehat{i}-\widehat{j}$

Hence, required dot product $= 6 - 1 = 5$