If the line x = y = z intersects the line x   sin A + y   sin B + z   sin C - 18 = 0 = x…

Given,

The line $x=y=z$ intersects the line $x \sin A+y \sin B+z \sin C-18=0=x \sin 2A+y \sin 2B+z\sin 2C-9$

So, let $x=y=z=k$

Now putting the value in $x \sin A+y \sin B+z \sin C=18$ we get,

$k\left(\sin A+\sin B+\sin C\right)=18$

Now we know that, if $A, B \& C$ are angles of triangle then $\sin A+\sin B+\sin C=4\cos \frac{A}{2}·\cos \frac{B}{2}·\cos \frac{C}{2}$

$\Rightarrow k\left(4\cos \frac{A}{2}·\cos \frac{B}{2}·\cos \frac{C}{2}\right)=18 .....\left(\mathrm{i}\right)$

Also $k\left(\sin 2A+\sin 2A+\sin 2A\right)=9$

And similalry $\sin 2A+\sin 2A+\sin 2A=4\sin A·\sin B·\sin C$

$\Rightarrow k\left(4\sin A·\sin B·\sin C\right)=9 ........\left(\mathrm{ii}\right)$

Now dividing equation $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$ we get,

$8 \sin \frac{A}{2}·\sin \frac{B}{2}·\sin \frac{C}{2}=\frac{9}{18}$

$\Rightarrow 80 \sin \frac{A}{2}·\sin \frac{B}{2}·\sin \frac{C}{2}=5$