If the distance between the planes 2x + y + z | vedclass

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(D) The given planes are $2x + y + z + 1 = 0$ and $2x + y + z + \alpha = 0$.Since the coefficients of $x, y, z$ are the same,the planes are parallel.T

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$-53$

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(D) The given planes are $2x + y + z + 1 = 0$ and $2x + y + z + \alpha = 0$.Since the coefficients of $x, y, z$ are the same,the planes are parallel.The distance $d$ between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$.Here,$A = 2, B = 1, C = 1, D_1 = 1, D_2 = \alpha$,and $d = 3$.Substituting these values,we get $3 = \frac{|1 - \alpha|}{\sqrt{2^2 + 1^2 + 1^2}}$.$3 = \frac{|1 - \alpha|}{\sqrt{4 + 1 + 1}} = \frac{|1 - \alpha|}{\sqrt{6}}$.$|1 - \alpha| = 3\sqrt{6}$.This implies $1 - \alpha = 3\sqrt{6}$ or $1 - \alpha = -3\sqrt{6}$.Thus,$\alpha = 1 - 3\sqrt{6}$ or $\alpha = 1 + 3\sqrt{6}$.The product of all possible values of $\alpha$ is $(1 - 3\sqrt{6})(1 + 3\sqrt{6})$.Using the identity $(a - b)(a + b) = a^2 - b^2$,we get $1^2 - (3\sqrt{6})^2 = 1 - (9 	imes 6) = 1 - 54 = -53$.

If the distance between the planes $2x + y + z + 1 = 0$ and $2x + y + z + \alpha = 0$ is $3$ units,then the product of all possible values of $\alpha$ is

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