Let the lines $\mathrm{L}_1: \vec{r}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}), \lambda \in \mathbb{R}$ and $\mathrm{L}_2: \vec{r}=(4 \hat{\mathrm{i}}+\hat{\mathrm{j}})+\mu(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$, intersect at the point R . Let P and Q be the points lying on lines $L_1$ and $L_2$, respectively, such that $|\overrightarrow{\mathrm{PR}}|=\sqrt{29}$ and $|\overrightarrow{\mathrm{PQ}}|=\sqrt{\frac{47}{3}}$. If the point P lies in the first octant, then $27(\mathrm{QR})^2$ is equal to

Let $\mathrm{S}=\left\{z \in \mathbb{C}:\left|\frac{z-6 i}{z-2 i}\right|=1\right.$ and $\left.\left|\frac{z-8+2 i}{z+2 i}\right|=\frac{3}{5}\right\}$.

Then $\sum\limits_{z \in \mathrm{~s}}|z|^2$ is equal to :

Consider an A.P.: $a_1, a_2, \ldots, a_{\mathrm{n}} ; a_1>0$. If $a_2-a_1=\frac{-3}{4}, a_{\mathrm{n}}=\frac{1}{4} a_1$, and $\sum\limits_{\mathrm{i}=1}^{\mathrm{n}} a_{\mathrm{i}}=\frac{525}{2}$, then $\sum\limits_{\mathrm{i}=1}^{17} a_{\mathrm{i}}$ is equal to

The mean and variance of a data of 10 observations are 10 and 2 , respectively. If an observations $\alpha$ in this data is replaced by $\beta$, then the mean and variance become 10.1 and 1.99 , respectively. Then $\alpha+\beta$ equals