For any real number k, if the line (3K + 2) x-ky-2 = 0 passes a certain point, then the fixed point

For the straight line (3K + 2) x-ky-2 = 0,
So (3x-y) K + 2 (x-1) = 0,
When X-1 = 0, 3x-y = 0, that is, x = 1, y = 3, for any real number k,
The equation is always true. A straight line must pass through points (1,3)

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