91Ó°ÊÓ

Firstly, note that in any triangle, the sum of the lengths of any two sides is always greater than the length of the remaining side. This means that: \[ a+b > c \]\[ b+c > a \]\[ a+c > b \]

Remember the given equation is \((a+b+c)(b+c-a)=k .(b c)\). If we take the middle term \((b+c-a)\) based on triangle inequalities, for example \(b+c\) is always greater than \(a\), then the term would be positive. Thus, the equation can be rewritten as: \[ (a+ b+ c) .x = k .(b .c) \] where \(x > 0\). The left side of this equation is always positive.

We must consider each condition for \( k \) separately: (a) If \( k<0 \), the equation \( (a+b+c) . x = k . (b c) \) is not possible, since both sides of the equation are positive. Thus, there are no valid triangles for this condition. (b) If \( k > 6 \), the equation \( (a+ b+ c) .x = k .(b .c) \) might hold, but only if the right side is larger than the left side. Here, since the parameter you weigh \(b.c\) with, is larger than 6, and a triangle's sides \(a, b, c\) need to fulfill \( a + b > c, b + c > a, a + c > b\), this cannot hold. (c) If \( 0 < k < 4 \), the equation \( (a+ b+ c) .x = k .(b .c) \) may hold in certain cases, creating valid triangles. (d) If \( k > 4 \), the equation \( (a + b + c) . x = k . (b c) \) might hold similar to case (b), but only if the right side is larger than the left side. Here, this cannot hold.