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(D) The equation of the plane passing through the line of intersection of $P_1$ and $P_2$ is given by $P_1 + kP_2 = 0$.$(x + (\lambda+4)y + z - 1) + k

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$3$

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(D) The equation of the plane passing through the line of intersection of $P_1$ and $P_2$ is given by $P_1 + kP_2 = 0$.$(x + (\lambda+4)y + z - 1) + k(2x + y + z - 2) = 0$  $(1)$Since the plane passes through $(0, 1, 0)$:$(0 + (\lambda+4)(1) + 0 - 1) + k(0 + 1 + 0 - 2) = 0$$\lambda + 3 - k = 0 \implies k = \lambda + 3$Since the plane passes through $(1, 0, 1)$:$(1 + 0 + 1 - 1) + k(2 + 0 + 1 - 2) = 0$$1 + k = 0 \implies k = -1$Equating the values of $k$:$\lambda + 3 = -1 \implies \lambda = -4$Now,the point is $(2\lambda, \lambda, -\lambda) = (-8, -4, 4)$.The distance of the point $(x_0, y_0, z_0)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.For plane $P_2: 2x + y + z - 2 = 0$ and point $(-8, -4, 4)$:$d = \frac{|2(-8) + 1(-4) + 1(4) - 2|}{\sqrt{2^2 + 1^2 + 1^2}} = \frac{|-16 - 4 + 4 - 2|}{\sqrt{6}} = \frac{|-18|}{\sqrt{6}} = \frac{18}{\sqrt{6}} = 3\sqrt{6}$.

Let the plane containing the line of intersection of the planes $P_1: x+(\lambda+4)y+z=1$ and $P_2: 2x+y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$. Then the distance of the point $(2\lambda, \lambda, -\lambda)$ from the plane $P_2$ is (in $\sqrt{6}$)

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