The points $A(2,1), B(3,-2)$ and $C(a, b)$ are vertices of the rectangle $A B C D$. If the point $P(3,4)$ lies on $C D$ produced, then $5 a+10 b=$

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, $5 a+10 b=5 \times \frac{24}{5}+10 \times\left(-\frac{7}{5}\right)=24-14=10$