(i) We need to find the value of $$\tan t$$ where $$t = \sum\limits_{i=1}^\infty \tan^{-1}\left(\frac{1}{2i^2}\right)$$.

We rewrite the sum: $$\tan^{-1}\left(\frac{1}{2i^2}\right)$$ can be written as $$\tan^{-1}\left(\frac{2}{(2i+1)-(2i-1)}\right)$$.

Using the formula: $$\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$$, we turn the sum into a telescoping series: $$\sum\limits_{i=1}^\infty [\tan^{-1}(2i+1) - \tan^{-1}(2i-1)]$$.

When you write out the terms, most of them cancel out. What is left is: $$t = \lim_{n \to \infty} [\tan^{-1}(2n+1) - \tan^{-1}(1)]$$.

As $$n$$ becomes very large, $$\tan^{-1}(2n+1)$$ approaches $$\frac{\pi}{2}$$, so: $$t = \frac{\pi}{2} - \tan^{-1}(1)$$.

We know that $$\tan^{-1}(1) = \frac{\pi}{4}$$, so: $$t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$.

So, $$\tan t = \tan \left(\frac{\pi}{4}\right) = 1$$.

(ii) The sides $$a, b, c$$ of a triangle are in arithmetic progression (AP). We have: $$\cos \theta_1 = \frac{a}{b+c}, \cos \theta_2 = \frac{b}{a+c}, \cos \theta_3 = \frac{c}{a+b}$$.

We know: $$\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$$.

So, $$\cos \theta_1 = \frac{1-\tan^2\left(\frac{\theta_1}{2}\right)}{1+\tan^2\left(\frac{\theta_1}{2}\right)} = \frac{a}{b+c}$$.

Using componendo and dividendo, we get: $$\tan^2\left(\frac{\theta_1}{2}\right) = \frac{b+c-a}{a+b+c}$$.

Similarly, $$\tan^2\left(\frac{\theta_3}{2}\right) = \frac{a+b-c}{a+b+c}$$.

So if we add them: $$\tan^2\left(\frac{\theta_1}{2}\right) + \tan^2\left(\frac{\theta_3}{2}\right) = \frac{b+c-a + a+b-c}{a+b+c} = \frac{2b}{a+b+c}$$.

If $$a,b,c$$ are in AP, then $$a + c = 2b$$, so $$a + b + c = 3b$$.

Therefore, $$\tan^2\left(\frac{\theta_1}{2}\right) + \tan^2\left(\frac{\theta_3}{2}\right) = \frac{2b}{3b} = \frac{2}{3}$$.

(iii) The line passes through (0, 1, 0) and is perpendicular to the plane $$x + 2y + 2z = 0$$.

The normal vector to the plane is (1, 2, 2). So, the direction vector for the line is (1, 2, 2).

The equation of the line is: $$\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-0}{2} = r$$.

Let $$P(r, 2r+1, 2r)$$ be a point on the line. To find the value of $$r$$ where the vector from the origin (0, 0, 0) to the point on the line is perpendicular to the direction vector (1, 2, 2), set up:

$$(r, 2r+1, 2r) \cdot (1, 2, 2) = 0$$

That is: $$r \times 1 + (2r+1)\times 2 + 2r \times 2 = 0$$

So: $$r + 4r + 2 + 4r = 0$$ $$9r + 2 = 0$$ $$r = -\frac{2}{9}$$

So the point is: $$\left(-\frac{2}{9}, \frac{5}{9}, -\frac{4}{9}\right)$$

The distance from the origin to this point is: $$\sqrt{\left(-\frac{2}{9}\right)^2 + \left(\frac{5}{9}\right)^2 + \left(-\frac{4}{9}\right)^2} = \sqrt{\frac{4+25+16}{81}} = \sqrt{\frac{45}{81}} = \frac{\sqrt{5}}{3}$$

(i)	$$\sum\limits_{i = 1}^\infty {{{\tan }^{ - 1}}\left( {{1 \over {2{i^2}}}} \right) = t} $$ then $$\tan t=$$	(A)	0
(ii)	Sides $$a,b,c$$ of a triangle ABC are in AP and $$\cos {\theta _1} = {a \over {b + c}},\cos {\theta _2} = {b \over {a + c}},\cos {\theta _3} = {c \over {a + b}}$$, then $${\tan ^2}\left( {{{{\theta _1}} \over 2}} \right) + {\tan ^2}\left( {{{{\theta _3}} \over 2}} \right) = $$	(B)	1
(iii)	A line is perpendicular to $$x + 2y + 2z = 0$$ and passes through (0, 1, 0). The perpendicular distance of this line from the origin is	(C)	$${{\sqrt 5 } \over 3}$$
		(D)	2/3

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