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Earlier discussion
1 older comment kept for context.
Victor Titus
For collinear points (points on the same line), the slopes are equal for any 2 points on the line.
Given (-1, t - 1), (t, t - 3), (t - 6, 3),
\(slope = \frac{(t-3) - (t-1)}{t - (-1)} = \frac{3 - (t-3)}{(t-6) - t} = \frac{3 - (t-1)}{(t-6) - (-1)}\)
Taking any two of the equations above, solve for t.
\(\frac{t - 3 - t + 1}{t + 1} = \frac{6 -t}{-6}\)
\(12 = (6 - t)(t + 1)\)
\(-t^{2} + 5t + 6 - 12 = 0 \implies t^{2} - 5t + 6 = 0\)
Solving, we have t = 2 and 3.