X and Y are the number of electrons involved, respectively during the oxidation of $\mathrm{I}^{-}$to $\mathrm{I}_2$ and $\mathrm{S}^{2-}$ to S by acidified $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$. The value of $\mathrm{X}+\mathrm{Y}$ is $\_\_\_\_$ .

Let $\vec{a}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\vec{c}=\vec{a} \times \vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11},|\vec{c} \times \vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a} \cdot \vec{d}$ is equal to

Let $A(1,0), B(2,-1)$ and $C\left(\frac{7}{3}, \frac{4}{3}\right)$ be three points. If the equation of the bisector of the angle ABC is $\alpha x+\beta y=5$, then the value of $\alpha^2+\beta^2$ is

Let $\mathrm{A}_1$ be the bounded area enclosed by the curves $y=x^2+2, x+y=8$ and $y$-axis that lies in the first quadrant. Let $\mathrm{A}_2$ be the bounded area enclosed by the curves $y=x^2+2, y^2=x, x=2$, and $y$-axis that lies in the first quadrant. Then $\mathrm{A}_1-\mathrm{A}_2$ is equal to