A BJT biasing circuit is shown in the figure, where V_BE = 0.7 V and β = 100. The Quiescent Point values of V_CE and I_C are respectively

The correct option is 4.6V and 2.46mA.

Based on the provided BJT biasing circuit diagram, the parameters are:
Supply voltage, $V_{\mathrm{CC}}=12\text{V}$
Base voltage-divider resistor, $R_1=100\text{k}\Omega$
Base voltage-divider resistor, $R_2=50\text{k}\Omega$
Collector resistor, $R_C=2\text{k}\Omega$
Emitter resistor, $R_E=1\text{k}\Omega$
Base-emitter voltage, $V_{\mathrm{BE}}=0.7\text{V}$
Common-emitter current gain, $\mathrm{��}=100$

Step 1: Calculate the Thevenin equivalent voltage ($V_{\mathrm{th}}$) and resistance ($R_{\mathrm{th}}$) at the base:
The Thevenin voltage is determined by the voltage divider network at the base node:
$$V_{\mathrm{th}}=V_{\mathrm{CC}}·\frac{R_2}{R_1+R_2}$$
Substituting the circuit values:

The Thevenin resistance is the parallel combination of the base resistors:
$$R_{\mathrm{th}}=\frac{R_1·R_2}{R_1+R_2}$$
Substituting the circuit values:
$$R_{\mathrm{th}}=\frac{100·50}{100+50}=\frac{5000}{150}\approx 33.33\text{k}\Omega$$

Step 2: Determine the base current ($I_B$):
Applying Kirchhoff's Voltage Law (KVL) to the base-emitter loop:
$$V_{\mathrm{th}}-I_B{R}_{\mathrm{th}}-V_{\mathrm{BE}}-I_E{R}_E=0$$
Since $I_E=(\beta +1)I_B$, we rewrite the relation to solve for $I_B$:
$$I_B=\frac{V_{\mathrm{th}}-V_{\mathrm{BE}}}{R_{\mathrm{th}}+(\beta +1)R_E}$$
Substituting the numerical parameters:
$$I_B=\frac{4-0.7}{33.33\text{k}\Omega +(100+1)·1\text{k}\Omega }=\frac{3.3}{134.33\text{k}\Omega }\approx 24.57\mu \text{A}$$

Step 3: Calculate the collector current ($I_C$):
Using the relationship between collector current and base current in the active region:
$$I_C=\beta ·I_B$$
$$I_C=100·24.57\mu \text{A}=2.457\text{mA}\approx 2.46\text{mA}$$

Step 4: Calculate the collector-emitter voltage ($V_{\mathrm{CE}}$):
First, find the emitter current ($I_E$):
$$I_E=(\beta +1)I_B=101·24.57\mu \text{A}\approx 2.48\text{mA}$$
Applying KVL around the collector-emitter output loop:
$$V_{\mathrm{CC}}-I_C{R}_C-V_{\mathrm{CE}}-I_E{R}_E=0$$
Rearranging to solve for $V_{\mathrm{CE}}$:
$$V_{\mathrm{CE}}=V_{\mathrm{CC}}-I_C{R}_C-I_E{R}_E$$
Substituting the parameters and calculated current values:
$$V_{\mathrm{CE}}=12-(2.46\text{mA}·2\text{k}\Omega )-(2.48\text{mA}·1\text{k}\Omega )$$
$$V_{\mathrm{CE}}=12-4.92-2.48=4.6\text{V}$$

Therefore, the Quiescent Point values are $V_{\mathrm{CE}}=4.6\text{V}$ and $I_C=2.46\text{mA}$.