A plane passes through a fixed point A(1, -2, | vedclass

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(D) Let the foot of the perpendicular from point $B(0, 1, 2)$ to the plane be $P(x, y, z)$.Since the plane passes through $A(1, -2, 3)$,the vector $\v

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$-1$

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(D) Let the foot of the perpendicular from point $B(0, 1, 2)$ to the plane be $P(x, y, z)$.Since the plane passes through $A(1, -2, 3)$,the vector $\vec{AP}$ lies in the plane.The vector $\vec{PB}$ is the normal to the plane at point $P$.Therefore,$\vec{AP} \cdot \vec{PB} = 0$.Given $A(1, -2, 3)$,$B(0, 1, 2)$,and $P(x, y, z)$:$\vec{AP} = (x-1, y+2, z-3)$$\vec{PB} = (0-x, 1-y, 2-z) = (-x, 1-y, 2-z)$Taking the dot product:$(x-1)(-x) + (y+2)(1-y) + (z-3)(2-z) = 0$$-x^2 + x + y - y^2 + 2 - 2y + 2z - z^2 - 6 + 3z = 0$$-x^2 - y^2 - z^2 + x - y + 5z - 4 = 0$Multiplying by $-1$:$x^2 + y^2 + z^2 - x + y - 5z + 4 = 0$Comparing this with $x^2 + y^2 + z^2 + \alpha x + \beta y + \gamma z + \delta = 0$,we get:$\alpha = -1, \beta = 1, \gamma = -5, \delta = 4$Thus,$\alpha + \beta + \gamma + \delta = -1 + 1 - 5 + 4 = -1$.

$A$ plane passes through a fixed point $A(1, -2, 3)$. If the locus of the foot of the perpendicular to it from the point $B(0, 1, 2)$ is $x^2 + y^2 + z^2 + \alpha x + \beta y + \gamma z + \delta = 0$,then the value of $\alpha + \beta + \gamma + \delta$ is

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