Let $\vec{a}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ and $\vec{b}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. If $\vec{c}$ is a vector such that $2(\vec{a} \times \vec{c})+3(\vec{b} \times \vec{c})=\overrightarrow{0}$ and $(\vec{a}-\vec{b}) \cdot \vec{c}=-97$, then $|\vec{c} \times \hat{\mathrm{k}}|^2$ is equal to

$\left(\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{3^2}+\frac{1}{3} \times \frac{4}{7}+\frac{4^2}{7^2}\right)+\left(\frac{1}{3^3}+\frac{1}{3^2} \times \frac{4}{7}+\frac{1}{3} \times \frac{4^2}{7^2}+\frac{4^3}{7^3}\right)+\ldots$ upto infinite terms, is equal to

Let S be a set of 5 elements and $\mathrm{P}(\mathrm{S})$ denote the power set of S . Let E be an event of choosing an ordered pair (A, B) from the set $\mathrm{P}(\mathrm{S}) \times \mathrm{P}(\mathrm{S})$ such that $\mathrm{A} \cap \mathrm{B}=\emptyset$. If the probability of the event $E$ is $\frac{3^p}{2^q}$, where $p, q \in N$, then $p+q$ is equal to

Let $z=(1+i)(1+2 i)(1+3 i) \ldots .(1+n i)$, where $i=\sqrt{-1}$. If $|z|^2=44200$, then $n$ is equal to $\_\_\_\_$