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(C) The equation of the plane passing through points $A(-1, -2, -3)$,$B(9, 3, 4)$,and $C(9, -2, 1)$ is given by the determinant:$\begin{vmatrix} x+1 &

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$\sqrt{42}$

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(C) The equation of the plane passing through points $A(-1, -2, -3)$,$B(9, 3, 4)$,and $C(9, -2, 1)$ is given by the determinant:$\begin{vmatrix} x+1 & y+2 & z+3 \ 9-(-1) & 3-(-2) & 4-(-3) \ 9-(-1) & -2-(-2) & 1-(-3) \end{vmatrix} = 0$$\begin{vmatrix} x+1 & y+2 & z+3 \ 10 & 5 & 7 \ 10 & 0 & 4 \end{vmatrix} = 0$Expanding along the third row:$10(7(y+2) - 5(z+3)) - 0 + 4(5(x+1) - 10(y+2)) = 0$$10(7y + 14 - 5z - 15) + 4(5x + 5 - 10y - 20) = 0$$10(7y - 5z - 1) + 4(5x - 10y - 15) = 0$$70y - 50z - 10 + 20x - 40y - 60 = 0$$20x + 30y - 50z - 70 = 0$Dividing by $10$,we get the plane equation: $2x + 3y - 5z - 7 = 0$.Let the foot of the perpendicular from $P(3, -2, -9)$ to the plane be $Q(x, y, z)$.The line passing through $P$ and perpendicular to the plane has direction ratios $(2, 3, -5)$.The equation of the line is $\frac{x-3}{2} = \frac{y+2}{3} = \frac{z+9}{-5} = k$.So,$x = 2k+3, y = 3k-2, z = -5k-9$.Since $Q$ lies on the plane,$2(2k+3) + 3(3k-2) - 5(-5k-9) - 7 = 0$.$4k + 6 + 9k - 6 + 25k + 45 - 7 = 0$$38k + 38 = 0 \implies k = -1$.Substituting $k = -1$,we get $Q(2(-1)+3, 3(-1)-2, -5(-1)-9) = Q(1, -5, -4)$.The distance of $Q(1, -5, -4)$ from the origin $(0, 0, 0)$ is $\sqrt{1^2 + (-5)^2 + (-4)^2} = \sqrt{1 + 25 + 16} = \sqrt{42}$.

Let the foot of the perpendicular from the point $P (3, -2, -9)$ on the plane passing through the points $A (-1, -2, -3)$,$B (9, 3, 4)$,and $C (9, -2, 1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is:

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