let y = -x² + ax - 6

dy/dx = -2x + a

For y = x - 2, dy/dx = 1

When line is a tangent to the curve, the gradient is the same for both at that point.

So -2x + a = 1

2x = a - 1

x = (a - 1)/2

Since the line is a tangent to the curve, they meet at one point.

So when -x² + ax - 6 = x - 2, there one real and repeated root .

x² + x - ax + 6 - 2 = 0
x² + (1 - a)x + 4 = 0

For one real and repeated root, discriminant = 0
(b² - 4ac = 0)

(1 - a)² - 4(1)(4) = 0

(1 - a)² - 16 = 0

(1 - a)² = 16

1 - a = ±√16 = ±4

a = 1 ± 4

a = -3 or a = 5

when a = -3,

x = (-3 - 1)/2 = -2

y = -2 - 2 = -4

Coordinates of P are (-2,-4)

Edited