Given $\quad \vec{A}=(2 \hat{i}-3 \hat{j}+\hat{k}), \quad \vec{B}=(3 \hat{i}+\hat{j}-2 \hat{k})$ and $\vec{C}=(3 \hat{i}+2 \hat{j}+\hat{k}) \cdot(\vec{A}+\vec{B}) \cdot \vec{C}$ will be

In the logic circuit given, $\mathrm{A}, \mathrm{B}$ and C are the inputs and Y is the output. The output Y is HIGH

At any time ' $t$ ', the co-ordinates of moving particle are $x=a t^2$ and $y=b t^2$. The speed of the particle is

A radioactive element A decays into radioactive element C by the following processes in succession.

$\mathrm{A} \rightarrow \mathrm{B}+{ }_2 \mathrm{H}_{\mathrm{e}}^4 ; \mathrm{B} \rightarrow \mathrm{C}+2 \mathrm{e}^{-}$Then elements