If the line x = y = z intersects the line xsinA + ysinB + zsinC -...

Given,

The line $x = y = z$ intersects the line $xsinA + ysinB + zsinC - 18 = 0 = xsin2A + ysin2B + zsin2C - 9$

So, let $x = y = z = k$

Now putting the value in $xsinA + ysinB + zsinC = 18$ we get,

$k\left(sinA + sinB + sinC\right) = 18$

Now we know that, if $A, B \text{ and } C$ are angles of triangle then $sinA + sinB + sinC = 4cos\dfrac{A}{2} \cdot cos\dfrac{B}{2} \cdot cos\dfrac{C}{2}$

$⇒ k\left(4cos\dfrac{A}{2} \cdot cos\dfrac{B}{2} \cdot cos\dfrac{C}{2}\right) = 18 . . . . . \left(i\right)$

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

And similalry $\sin 2A + \sin 2A + \sin 2A = 4sinA \cdot sinB \cdot sinC$

$⇒ k\left(4sinA \cdot sinB \cdot sinC\right) = 9 . . . . . . . . \left(ii\right)$

Now dividing equation \left(i\right) \text{ and } \left(ii\right)$we get,</p><p>$8sin\dfrac{A}{2} \cdot sin\dfrac{B}{2} \cdot sin\dfrac{C}{2} = \dfrac{9}{18}$</p><p>$⇒ 80sin\dfrac{A}{2} \cdot sin\dfrac{B}{2} \cdot sin\dfrac{C}{2} = 5