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(D) The distance of the point $(1, 1, 1)$ from the origin $(0, 0, 0)$ is given by $\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.The distance of the point $(1, 1

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$3, -9$

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(D) The distance of the point $(1, 1, 1)$ from the origin $(0, 0, 0)$ is given by $\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.The distance of the point $(1, 1, 1)$ from the plane $x + y + z + k = 0$ is given by $\frac{|1 + 1 + 1 + k|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|k + 3|}{\sqrt{3}}$.According to the problem,the distance from the origin is half the distance from the plane:$\sqrt{3} = \frac{1}{2} 	imes \frac{|k + 3|}{\sqrt{3}}$.Multiplying both sides by $2\sqrt{3}$,we get:$2 	imes 3 = |k + 3|$$|k + 3| = 6$.This implies $k + 3 = 6$ or $k + 3 = -6$.Solving these,we get $k = 3$ or $k = -9$.

If the distance of the point $(1, 1, 1)$ from the origin is half its distance from the plane $x + y + z + k = 0$,then $k = $

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