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

The equation of the plane containing their line of intersection is:

$$\left[ x + (\lambda + 4)y + z - 1 \right] + k \left[ 2x + y + z - 2 \right] = 0 \quad \text{--- (Equation 1)}$$

$$\left[ 0 + (\lambda + 4)(1) + 0 - 1 \right] + k \left[ 2(0) + 1 + 0 - 2 \right] = 0$$

$$(\lambda + 3) + k(-1) = 0 \implies k = \lambda + 3 \quad \text{--- (Equation 2)}$$

$$\left[ 1 + (\lambda + 4)(0) + 1 - 1 \right] + k \left[ 2(1) + 0 + 1 - 2 \right] = 0$$

$$1 + k(1) = 0 \implies k = -1$$

$$\lambda + 3 = -1 \implies \lambda = -4$$

$$\text{Point } = (2(-4), \, -4, \, -(-4)) = (-8, \, -4, \, 4)$$

The equation of plane $$P_2$$ is: $$2x + y + z - 2 = 0$$

Using the distance formula for the point $$(-8, -4, 4)$$: $$d = \frac{|2(-8) + 1(-4) + 1(4) - 2|}{\sqrt{2^2 + 1^2 + 1^2}}$$

$$d = \frac{|-16 - 4 + 4 - 2|}{\sqrt{4 + 1 + 1}}$$

$$d = \frac{|-18|}{\sqrt{6}} = \frac{18}{\sqrt{6}}$$

$$d = \frac{18\sqrt{6}}{6} = 3\sqrt{6}$$