If the plane -4x - 2y + 2z + alpha = 0 is at | vedclass

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(C) The given planes are $P_1: -4x - 2y + 2z + \alpha = 0$ and $P_2: 2x + y - z + 1 = 0$. First,rewrite $P_1$ by dividing by $-2$: $2x + y - z - \frac

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-$92$

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(C) The given planes are $P_1: -4x - 2y + 2z + \alpha = 0$ and $P_2: 2x + y - z + 1 = 0$. First,rewrite $P_1$ by dividing by $-2$: $2x + y - z - \frac{\alpha}{2} = 0$. Let $k = -\frac{\alpha}{2}$. The planes are $2x + y - z + k = 0$ and $2x + y - z + 1 = 0$. 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,$d = 2$,$A = 2$,$B = 1$,$C = -1$,$D_1 = k$,and $D_2 = 1$. $2 = \frac{|k - 1|}{\sqrt{2^2 + 1^2 + (-1)^2}} = \frac{|k - 1|}{\sqrt{6}}$. $|k - 1| = 2\sqrt{6}$. $k - 1 = 2\sqrt{6}$ or $k - 1 = -2\sqrt{6}$. $k = 1 + 2\sqrt{6}$ or $k = 1 - 2\sqrt{6}$. Since $k = -\frac{\alpha}{2}$,we have $\alpha = -2k$. $\alpha_1 = -2(1 + 2\sqrt{6}) = -2 - 4\sqrt{6}$ and $\alpha_2 = -2(1 - 2\sqrt{6}) = -2 + 4\sqrt{6}$. The product of the values of $\alpha$ is $\alpha_1 \alpha_2 = (-2 - 4\sqrt{6})(-2 + 4\sqrt{6}) = (-2)^2 - (4\sqrt{6})^2 = 4 - 16(6) = 4 - 96 = -92$.

If the plane $-4x - 2y + 2z + \alpha = 0$ is at a distance of $2$ units from the plane $2x + y - z + 1 = 0$,then the product of all the possible values of $\alpha$ is

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