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(D) Since $C(0, -a, -1)$ lies on the plane $lx + my + nz = 0,$ we have $l(0) + m(-a) + n(-1) = 0,$ which implies $-ma - n = 0,$ or $\frac{m}{n} = -\fr

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$\sqrt{66}$

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(D) Since $C(0, -a, -1)$ lies on the plane $lx + my + nz = 0,$ we have $l(0) + m(-a) + n(-1) = 0,$ which implies $-ma - n = 0,$ or $\frac{m}{n} = -\frac{1}{a} \quad \dots(1)$The vector $\vec{CA} = (a - 0, -2a - (-a), 3 - (-1)) = (a, -a, 4)$ is parallel to the normal vector of the plane $\vec{n} = (l, m, n).$Thus,$\frac{a}{l} = \frac{-a}{m} = \frac{4}{n}.$ From $\frac{-a}{m} = \frac{4}{n},$ we get $\frac{m}{n} = -\frac{a}{4} \quad \dots(2)$Equating $(1)$ and $(2),$ we have $-\frac{1}{a} = -\frac{a}{4} \Rightarrow a^2 = 4.$ Since $a > 0,$ we have $a = 2.$Substituting $a = 2$ into $(1),$ we get $\frac{m}{n} = -\frac{1}{2}.$ Let $m = -t$ and $n = 2t.$ Then $\frac{2}{l} = \frac{-2}{-t} \Rightarrow l = t.$The equation of the plane is $t(x - y + 2z) = 0,$ or $x - y + 2z = 0.$The point $D$ is the foot of the perpendicular from $B(0, 4, 5)$ to the plane $x - y + 2z = 0.$ The line $BD$ is given by $\frac{x-0}{1} = \frac{y-4}{-1} = \frac{z-5}{2} = k.$So,$D = (k, 4-k, 5+2k).$ Since $D$ lies on the plane,$k - (4-k) + 2(5+2k) = 0 \Rightarrow k - 4 + k + 10 + 4k = 0 \Rightarrow 6k + 6 = 0 \Rightarrow k = -1.$Thus,$D = (-1, 4-(-1), 5+2(-1)) = (-1, 5, 3).$With $C = (0, -2, -1)$ and $D = (-1, 5, 3),$ the length $CD = \sqrt{(-1-0)^2 + (5-(-2))^2 + (3-(-1))^2} = \sqrt{(-1)^2 + 7^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66}.$

If for $a > 0,$ the feet of perpendiculars from the points $A(a, -2a, 3)$ and $B(0, 4, 5)$ on the plane $lx + my + nz = 0$ are points $C(0, -a, -1)$ and $D$ respectively,then the length of line segment $CD$ is equal to

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