A) 2x ? y ? z ? 1 = 0
B) 2x + 3y + z ? 1 = 0
C) 2x + y + z + 3 = 0
D) x ? y + z ? 1 = 0
Correct Answer: A
Solution :
Equation of plane passes through (2, 2, 1) is \[{{H}_{2}}\left( g \right)={{p}_{2}}mm\] Since, above plane is perpendicular to 3x + 2y + 4z + 1 = 0 and 2x + y + 3z + 2 = 0 \[\therefore \] 3a + 2b + 4c = 0 and 2a + b + 3c = 0 \[{{p}_{1}}+{{p}_{2}}=60mm\]On multiplying Eq. (iii) by 2, we get 4a + 2b + 6c = 0 On subtracting Eq. (iv) from Eq. (ii), we get -a ? 2c = 0 \[{{C}_{6}}{{H}_{6}}\left( g \right)=0\] \[{{H}_{2}}\left( g \right)={{p}_{2}}-3{{p}_{1}}\] \[{{C}_{6}}{{H}_{12}}\left( g \right)={{p}_{1}}\] \[={{p}_{2}}-3{{p}_{1}}+{{p}_{1}}=30mm\] On putting \[{{p}_{2}}-2{{p}_{1}}=30mm\] in Eq. (iii), we get \[{{\text{p}}_{\text{1}}}\text{=10mm, }{{\text{p}}_{\text{2}}}\text{=50mm}\] \[{{C}_{6}}{{H}_{6}}\] 4a + 2b ? 3a = 0 \[=\frac{10}{60}=\frac{1}{6}\] 2b = - a \[=8\times \frac{1}{8}+6\times \frac{1}{2}=4\] \[=12\times \frac{1}{4}=3\] On putting \[\text{C}{{\text{u}}_{\text{4}}}\text{A}{{\text{g}}_{\text{3}}}\text{Au}\text{.}\] and \[-nF{{E}^{\circ }}_{cell}=-RT\] in Eq. (i). We get \[{{E}^{\circ }}_{cell}=\frac{RT}{nF}\] \[{{E}^{\circ }}_{cell}\] \[=\frac{1}{2}\frac{RT}{F}\] \[\begin{align} & N{{H}_{4}}Cl\xrightarrow{\Delta }N{{H}_{3}}+HCl \\ & ABC \\ \end{align}\] 2x ? 4 ? y + 2 ? z + 1 = 0 \[\begin{align} & N{{H}_{3}}+HCl\to N{{H}_{4}}Cl \\ & \text{Bwhite fumes} \\ \end{align}\] 2x ? y ? z ? 1 = 0
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