(3)

Let the line makes angle $\alpha$ with positive x-axis, then $\beta =\frac{\alpha }{2}$ and $\gamma =\frac{\alpha }{2}$

Now, ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

$\Rightarrow {\cos }^2\alpha +2{\cos }^2\frac{\alpha }{2}=1$

$\Rightarrow {\cos }^2\alpha +\cos \alpha =0$

$\Rightarrow \cos \alpha (\cos \alpha +1)=0$

$\Rightarrow \cos \alpha =0,–1\Rightarrow \alpha =\frac{\pi }{2},\pi$

Now, $\beta =\frac{\alpha }{2} \Rightarrow \beta =\frac{\pi }{4},\frac{\pi }{2}$

So, required sum $=\frac{\pi }{4}+\frac{\pi }{2}=\frac{3\pi }{4}$.

(1)

Given, A(x, y, z) be a point in xy-plane. Let the point P(0, 3, 2), Q(2, 0, 3) and R(0, 0, 1)

The distance of the point AP = AQ = AR

$\Rightarrow {(x–0)}^2+{(y–3)}^2+{(z–2)}^2$

     $={(x–2)}^2+{(y–0)}^2+{(z–3)}^2$

     $={(x–0)}^2+{(y–0)}^2+{(z–1)}^2$

In xy-plane, z = 0

So, $x^2–4x+y^2+9+4=x^2+y^2+1 \Rightarrow –4x+13=1 \Rightarrow x=3$

And $x^2+y^2–6y+9+4=x^2+y^2+1 \Rightarrow y=2$

So, A(3, 2, 0), B(1, 4, –1) and C(2, 0, –2).

In $△$ABC

$AB=\sqrt{1+4+4}=3, BC=\sqrt{1+16+1}=\sqrt{18}, CA=\sqrt{1+4+4}=3$

So, AB = AC and $A{B}^2+A{C}^2={\left(BC\right)}^2$

$\therefore$   $△$ABC is an isosceles right angled triangle.

So, (S1) is true.

Also, Area of $△ABC=\frac{1}{2}\times \left(\mathrm{Base}\right)\left(\mathrm{Height}\right)=\frac{1}{2}\times 3\times 3=\frac{9}{2}$

So, (S2) is false.