Option 2 : 13 - z^{2}

**Given :**

3cosθ + 2sinθ = z

**Formula used :**

If, ax + by = m, ay - bx = n

then, (a^{2} + b^{2})(x^{2} + y^{2}) = m^{2} + n^{2}

**Calculations :**

Let, 3sinθ - 2cosθ be x

Now,

3cosθ + 2sinθ = z ----(1)

3sinθ - 2cosθ = x ----(2)

Comparing with above given formula

(3^{2} + 2^{2}) (sin^{2}θ + cos^{2}θ) = z^{2 }+ x^{2 }

⇒ 13 × 1 = z2 + x2

⇒ x^{2} = 13 - z^{2}

So,

(3sinθ - 2cosθ)2 = x^{2}

⇒ 13 - z^{2}

**∴ The value of (3sinθ - 2cosθ) ^{2} is 13 - z^{2}**