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(B) The family of planes passing through the line of intersection of the planes $P_1: x + y + z - 5 = 0$ and $P_2: 2x + 3y + 4z + 5 = 0$ is given by $

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$x - z = 20$

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(B) The family of planes passing through the line of intersection of the planes $P_1: x + y + z - 5 = 0$ and $P_2: 2x + 3y + 4z + 5 = 0$ is given by $P_2 + \lambda P_1 = 0$.$(2x + 3y + 4z + 5) + \lambda(x + y + z - 5) = 0$$(2 + \lambda)x + (3 + \lambda)y + (4 + \lambda)z + (5 - 5\lambda) = 0$Since this plane is perpendicular to the plane $x + y + z = 5$,the dot product of their normal vectors must be zero.The normal vector of the required plane is $\vec{n_1} = ((2 + \lambda), (3 + \lambda), (4 + \lambda))$ and the normal vector of the given plane is $\vec{n_2} = (1, 1, 1)$.$(2 + \lambda)(1) + (3 + \lambda)(1) + (4 + \lambda)(1) = 0$$9 + 3\lambda = 0 \Rightarrow \lambda = -3$.Substituting $\lambda = -3$ into the family equation:$(2 - 3)x + (3 - 3)y + (4 - 3)z + (5 - 5(-3)) = 0$$-x + 0y + z + 20 = 0$$x - z = 20$.

The equation of the plane passing through the line of intersection of the planes $x + y + z = 5$ and $2x + 3y + 4z + 5 = 0$ and perpendicular to the plane $x + y + z = 5$ is

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