Given, $A B C D$ is a rectangle


Slope of $A B=\frac{-2-1}{3-2}=-3=m_1$ $\ldots(\mathrm{i})$
Slope of $B C=\frac{b+2}{a-3}=m_2$ $\ldots(\mathrm{ii})$
Now $\quad A B \perp B C \Rightarrow m_1 m_2=-1$ $\ldots(\mathrm{iii})$
$\therefore$ Putting value of $m_1$ and $m_2$ in Eq. (iii)
$-3\left(\frac{b+2}{a-3}\right)=-1$
$-3 b-6=-a+3 \Rightarrow a-3 b-9=0 \ldots$ (iv)
Now, slope of line $C D=$ slope of $C P$
$m_3=\frac{b-4}{a-3}$
$\therefore$ line $\quad A B \| C D \Rightarrow m_1=m_3 \quad \therefore-3=\frac{b-4}{a-3}$
$-3 a+9=b-4 \Rightarrow 3 a+b-13=0 \quad \ldots(\mathrm{v})$
Now, solving Eqs. (iv) and (v),
$a-3 b-9=0 \Rightarrow 3 a+b-13=0$
Eq. (iv) $\times(3)$ and subtracting Eq. (iv) and Eq. (v),

$\Rightarrow \quad b=\frac{-14}{10} \Rightarrow b=\frac{-7}{5}$
These value is putting in Eq. (iv)
$a-3\left(-\frac{7}{5}\right)-9=0 \Rightarrow a+\frac{21}{5}-9=0$
$a+\left(\frac{24}{5}\right)=0 \Rightarrow a=\frac{24}{5}$
Now, $5 a+10 b=5 \times \frac{24}{5}+10 \times\left(-\frac{7}{5}\right)=24-14=10$