We are given two unit vectors $$\vec{a}$$ and $$\vec{b}$$ with an angle of $$\frac{\pi}{3}$$ between them. This means:

$$\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = 1$$ (since they are unit vectors)

$$\vec{a} \cdot \vec{b} = \cos\frac{\pi}{3} = \frac{1}{2}$$

We need to find the number of values of $$\lambda$$ in the interval $$[-1, 3]$$ for which the vectors $$\lambda \vec{a}+2 \vec{b}$$ and $$3 \vec{a}-\lambda \vec{b}$$ are perpendicular, meaning their dot product is zero.

Here’s how to solve the problem step by step:

Set up the perpendicular condition by equating the dot product to zero:

$$(\lambda \vec{a}+2 \vec{b})\cdot(3\vec{a}-\lambda \vec{b}) = 0.$$

Expand the dot product using the distributive property:

$$\begin{align*} (\lambda \vec{a}+2 \vec{b})\cdot(3\vec{a}-\lambda \vec{b}) &= \lambda \cdot 3 (\vec{a}\cdot\vec{a}) - \lambda^2 (\vec{a}\cdot\vec{b}) + 2\cdot 3 (\vec{b}\cdot\vec{a}) - 2\lambda (\vec{b}\cdot\vec{b})\\[1mm] &= 3\lambda - \lambda^2\left(\frac{1}{2}\right) + 6\left(\frac{1}{2}\right) - 2\lambda\\[1mm] &= 3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda. \end{align*}$$

Simplify the expression:

$$3\lambda - 2\lambda = \lambda,$$

so the equation becomes:

$$\lambda - \frac{\lambda^2}{2} + 3 = 0.$$

Multiply the entire equation by 2 to eliminate the fraction:

$$2\lambda - \lambda^2 + 6 = 0,$$

which can be rearranged as:

$$-\lambda^2 + 2\lambda + 6 = 0.$$

Multiply both sides by -1 to get a standard quadratic form:

$$\lambda^2 - 2\lambda - 6 = 0.$$

Solve the quadratic equation using the quadratic formula:

$$\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4\cdot1\cdot(-6)}}{2} = \frac{2 \pm \sqrt{4+24}}{2} = \frac{2 \pm \sqrt{28}}{2}.$$

Notice that:

$$\sqrt{28} = 2\sqrt{7},$$

thus:

$$\lambda = \frac{2 \pm 2\sqrt{7}}{2} = 1 \pm \sqrt{7}.$$

Evaluate the solutions numerically:

$$\lambda = 1 + \sqrt{7} \approx 1 + 2.6458 \approx 3.6458,$$

$$\lambda = 1 - \sqrt{7} \approx 1 - 2.6458 \approx -1.6458.$$

Check if the solutions are in the interval $$[-1, 3]$$:

$$1 + \sqrt{7} \approx 3.6458$$ is greater than 3.

$$1 - \sqrt{7} \approx -1.6458$$ is less than -1.

Since neither value lies within the interval $$[-1, 3]$$, there are no valid values of $$\lambda$$ in that interval.

Therefore, the number of values of $$\lambda$$ in the interval $$[-1, 3]$$ is:

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