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(C) The equation of a plane passing through the line of intersection of the planes $P_1: x + 2y - 3 = 0$ and $P_2: y - 2z + 1 = 0$ is given by $P_1 +

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$2x - y + 10z = 11$

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(C) The equation of a plane passing through the line of intersection of the planes $P_1: x + 2y - 3 = 0$ and $P_2: y - 2z + 1 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(x + 2y - 3) + \lambda(y - 2z + 1) = 0$$x + (2 + \lambda)y - 2\lambda z + (\lambda - 3) = 0$  ....$(i)$Since this plane is perpendicular to the plane $x + 2y - 3 = 0$,the dot product of their normal vectors $\vec{n_1} = (1, 2 + \lambda, -2\lambda)$ and $\vec{n_2} = (1, 2, 0)$ must be zero.$1(1) + 2(2 + \lambda) + 0(-2\lambda) = 0$$1 + 4 + 2\lambda = 0$$5 + 2\lambda = 0 \Rightarrow \lambda = -\frac{5}{2}$Substituting $\lambda = -\frac{5}{2}$ into equation $(i)$:$x + (2 - \frac{5}{2})y - 2(-\frac{5}{2})z + (-\frac{5}{2} - 3) = 0$$x - \frac{1}{2}y + 5z - \frac{11}{2} = 0$Multiplying by $2$:$2x - y + 10z - 11 = 0$$2x - y + 10z = 11$

The equation of a plane through the line of intersection of the planes $x + 2y = 3$ and $y - 2z + 1 = 0$,and perpendicular to the first plane $x + 2y = 3$ is:

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