Any plane through (1, 1, –1 ) isa(x – 1) + b(y – 1) + c(z + 1) = 0 ...(1)Since the plane passes through the points (6, 4, –5) and (–4, –2, 3)∴ a(6 – 1) + b(4 – 1) + c (–5 + 1) = 0 and a(– 4 – 1) + b(–2 – 1) + c ( 3 + 1) = 0⇒ 5a + 3b – 4c = 0 ...(2)and –5a – 3b + 4c = 0 ...(3)Solving (2) and (3), $\frac{a}{12-12}=\frac{b}{20-20}=\frac{c}{-15+15}$$\Rightarrow\frac{a}{0}=\frac{b}{0}=\frac{c}{0}$⇒ a, b, c can’t be found.Since the points are collinear, so an infinite number of planes can be found through the given points.