$[3327]$

IN THE HIGH COURT OF ANDHRA PRADESH AT AMARAVATI

WEDNESDAY, THE THIRTIETH DAY OF NOVEMBER TWO THOUSAND AND TWENTY TWO

:PRESENT: THE HONOURABLE SRI JUSTICE K SREENIVASA REDDY

IA No. 3 OF 2022 IN CRLRC NO: 1111 OF 2022

Between:

State rep by Inspector of Police, CBI, ACB, Visakhapatnam Through its Assistant Solicitor General, High Court of Andhra Pradesh.

Petitioner

AND

Shri Bukya Balu, S/o Shri Hariya, Age. 42 years, Village Jamalapuram, Errupalem Mandal, Khammam District, Telangana State (Then Andhra Pradesh).

Respondent

Sd/- M. SRINIVAS

SECTION OFFICER

ASSISTANT REGISTRAR

$\blacktriangle$

Counsel for the Petitioner :HARINATH N (Asst Solicitor General) Counsel for the Respondent :

Petition under Section 482 of CrPC praying that in the circumstances stated in the affidavit filed in support of the petition, the High Court may be pleased to stay all further proceedings in the main CC. 16/2011 on the file of II Addl. Special Judge for CBI cases, Visakhapatnam and thus render justice, Pending disposal of CRLRC No. 1111 of 2022, on the file of the High Court.

The court while directing issue of notice to the Respondents herein to show cause as to why this application should not be complied with, made the following order.(The receipt of this order will be deemed to be the receipt of notice in the case). The Court made the following but.

ORDER:

"In the circumstances, there shall be an interim stay of all further proceedings in CC No.16 of 2011 on the file of the learned II Additional Special Judge for CBI Cases, Visakhapatnam for a period of six (6) weeks."

//TRUE COPY//

$\mathcal{L}^{\mathcal{L}}{\mathcal{L}}(\mathcal{L}) \subset \mathcal{M}^{\mathcal{L}}{\mathcal{L}}(\mathcal{L})$

$\mathcal{L} \cdot \mathcal{L} \cdot \mathcal{L}$

이 : 의 뉴.

To,

1. The II Addl. Special Judge for CBI cases, Visakhapatnam

• 2. Shri Bukya Balu, S/o Shri Hariya, Age. 42 years, Village Jamalapuram, Errupalem Mandal, Khammam District, Telangana State (Then Andhra Pradesh). (By RPAD) $\mathcal{L} = \mathcal{L} = \frac{\partial \hat{D}}{\partial \omega} \mathcal{L}$
• 3. One CC to SRI. HARINATH N (Asst Solicitor General) Advocate [OPUC]<br>4. One spare copy

经信息 经不值额的证据

$\label{eq:1} \begin{split} \mathcal{L}{\text{max}} = \frac{1}{\sqrt{2}} \sum{\mathbf{k}} \frac{1}{\sqrt{2}} \frac{\mathbf{k} \cdot \mathbf{r}^2}{\mathbf{k}^2} \frac{\mathbf{k} \cdot \mathbf{r}^2}{\mathbf{k}^2} \frac{\mathbf{k} \cdot \mathbf{r}^2}{\mathbf{k}^2} \frac{\mathbf{k} \cdot \mathbf{r}^2}{\mathbf{k}^2} \frac{\mathbf{k} \cdot \mathbf{r}^2}{\mathbf{k}^2} \frac{\mathbf{k} \cdot \mathbf{r}^2}{\mathbf{k}^2} \frac{\mathbf{k} \cdot \mathbf$ $\mathcal{L}^{\mathcal{A}}{\mathcal{A}}\left(\mathcal{A}\right)=\mathcal{L}^{\mathcal{A}}{\mathcal{A}}\left(\mathcal{A}\right)$

$\mathcal{L}^{\mathcal{L}}_{\mathcal{L}}\left(\mathcal{L}\right)$ $\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{$

$\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}$ $\mathcal{L}{\mathcal{L}}$ $\mathcal{L}^{\mathcal{L}}{\mathcal{L}}$

$\mathcal{H}^{\text{max}}{\text{max}}(\mathcal{H}^{\text{max}}{\text{max}})$ $\mathcal{O}(\sqrt{1-\frac{1}{2}}\log n)$

$\beta_{\rm{max}} = \beta_{\rm{max}} \left( \theta_{\rm{max}}^{\rm{max}} \right)$ $C_{\mathcal{M}}\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)^{2}$

$\label{eq:1} \mathcal{L}{\text{max}}\left(\mathcal{L}{\text{max}}\right) = \frac{1}{\sqrt{2}}\left(\mathcal{L}{\text{max}}\right)\mathcal{L}{\text{max}}\left(\mathcal{L}{\text{max}}\right) + \frac{1}{\sqrt{2}}\mathcal{L}{\text{max}}\left(\mathcal{L}_{\text{max}}\right)$

$\mathcal{L}{\mathcal{A}} = \mathcal{L}{\mathcal{A}} \otimes \mathcal{L}_{\mathcal{A}}$ $\mathbb{E}[\mathbf{x}^{\top},\mathbf{x}^{\top},\mathbf{x}]$

$\gamma(\kappa) = \gamma(\kappa)$ $\mathcal{L}^{(n)}$

$\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\right)^{1/2}$

$\mathcal{C}^{\mathcal{A}}_{\mathcal{A}}(x)$

$\hat{\mathcal{F}}$

${f(x)}_{x\in\mathbb{R}}\in\mathbb{R}^{n\times n}$

$\mathcal{A}^{\pm}$

$\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\right)^{1/2} \left(\frac{1}{\sqrt{2}}\right)^{1/2} \left(\frac{1}{\sqrt{2}}\right)^{1/2}$ $\langle \cdot \rangle_{\mathcal{C}}$

$\mathcal{A} = \mathcal{A} \cup \mathcal{A}$ $\begin{array}{c} \mathbf{y} \ \mathbf{y} \ \mathbf{y} \end{array} \rightarrow \begin{array}{c} \mathbf{y} \ \mathbf{y} \ \mathbf{y} \end{array}$ $\mathcal{L}^{\mathcal{A}}{\mathcal{A}}\subset\mathcal{L}^{\mathcal{A}}{\mathcal{A}}$ $\frac{d\mathcal{L}}{d\mathcal{L}}\sum_{i=1}^n\frac{d\mathcal{L}}{d\mathcal{L}}$

$\sqrt{2}(\mathcal{R})\sqrt{2}(\mathcal{R})\geq 0$

$\mathcal{L}^{\mathcal{L}}_{\mathcal{L}}(\mathcal{L})$

$\gamma\in\mathcal{C}_{\mathbb{R}^d}(\mathbb{R})$

$\mathcal{L}(\mathcal{L})$

$\frac{1}{\sqrt{2}}$

$\mathcal{L} = \begin{pmatrix} \mathcal{L} & \mathcal{L} \ \mathcal{L} & \mathcal{L} \end{pmatrix} \mathcal{L} \begin{pmatrix} \mathcal{L} \ \mathcal{L} \end{pmatrix}$

4. One spare copy

$ER$

$\sim$

$\mathfrak{f}$

$\overline{a}$

$\mathbb{E}[\mathcal{F}{\mathcal{A}}(\mathcal{A})]=\mathcal{A}{\mathcal{A}}$ $\mathcal{A}^{\mathcal{A}}$

HIGH COURT

$\text{SRK},,\text{J}$

DATED:30/11/2022

ORDER IA NO.3 OF 2022 IN CRLRC.No.1111 of 2022

STAY

$\overline{1}$