In an equilateral triangle of side length 6 , pegs are placed at...

Case I:
To divide the triangle into 9 regions. Two pegs must be selected from a side and the other two from a different side. This can be done in ${ }^3 C_2 \times{ }^5 C_2 \times{ }^5 C_2=300$ ways Case II:
Now we are choosing 3 points on three sides, such that three lines from those points are concurrent. By using Ceva's theorem in which product of three different ratio leads to 1 . Possible ratio on side $A B, B C$ and $C A$ will be of the form $\frac{m}{n}, \frac{n}{m}$ and 1 i.e. $\frac{m}{n} \times \frac{n}{m} \times 1=1$ Ratio $1: 1$ can be choosen in 3 ways for all three sides other ratio can be choosen in 4 ways for other two sides i.e. there are $3 \times 4+1=13$ ways Fourth point can be choosen in $12_{C_1}$, ways Total such possibilities $=12 \times 13=156$ ways Total ways $=300+156=456$ Sum of squares of digit $=4^2+5^2+6^2=16+25+36=77$