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(C) The normal vector $\vec{n}$ to the plane is the vector $\vec{PF}$.$\vec{PF} = (1-2, -2-0, 0-(-3)) = (-1, -2, 3)$.Since the equation of the plane i

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$6$

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(C) The normal vector $\vec{n}$ to the plane is the vector $\vec{PF}$.$\vec{PF} = (1-2, -2-0, 0-(-3)) = (-1, -2, 3)$.Since the equation of the plane is $ax+by-3z+d=0$,the normal vector is $(a, b, -3)$.Comparing $(a, b, -3)$ with $k(-1, -2, 3)$,we get $k(-1) = a$,$k(-2) = b$,and $k(3) = -3$.Thus,$k = -1$.Therefore,$a = -1(-1) = 1$ and $b = -1(-2) = 2$.The equation of the plane is $x+2y-3z+d=0$.Since the point $F(1,-2,0)$ lies on the plane,we have $1 + 2(-2) - 3(0) + d = 0$.$1 - 4 + d = 0 \implies -3 + d = 0 \implies d = 3$.We need to find $a+b+d = 1 + 2 + 3 = 6$.

If the foot of the perpendicular drawn from the point $P(2,0,-3)$ to the plane $\pi$ is $F(1,-2,0)$ and the equation of the plane $\pi$ is $ax+by-3z+d=0$,then $a+b+d=$

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