The equation of the given line is \( \sqrt{3}x+y+2=0\)

This equation can be reduced as,

\( -\sqrt{3}x-y=2\)

Divide both sides by

\(\sqrt{(-\sqrt{3})^2+(-1)^2}=2\),

We get,

\( \dfrac{-\sqrt{3}}{2}x-\dfrac{1}{2}y=\dfrac{2}{2}\)

\( \dfrac{-\sqrt{3}}{2}x-\dfrac{1}{2}y=1\) ...........(1)

Compare equation (1) to xcos θ+ ysin θ= p , we get

cos θ = \( \dfrac{-\sqrt{3}}{2}\), sin θ = \(- \dfrac{1}{2}\), and p=1

Therefore, the respective values of θ and p are

\( \dfrac{7\pi}{6}\) and 1.

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