determining k

For the three points to be collinear, the following determinant must be zero, whose elements are,

First row:       5 2 1

Second row:  k 3 1

Third row:      G k 1

5(3-k)-2(k-G)+(k2-3G) = 0

k² -7k +15-G = 0

k = [7±(49-4(15-G))]/2

= [7±(4G - 11)]/2

Regards,

Team,

TopperLearning.