Let $$\overrightarrow{a}$$ and $$ \overrightarrow{b}$$ be two unit vectors such that the angle between them is $$\frac{\pi}{3}$$. Tf $$\lambda \overrightarrow{a} +2\overrightarrow{b}\text{ and }3\overrightarrow{a}-\lambda \overrightarrow{b}$$ are perpendicular to each other, then the number of values of $$\lambda$$ in [-1,3] is :

Let $$\vec{a}$$ and $$\vec{b}$$ be unit vectors with the angle $$\frac{\pi}{3}$$ between them, and consider the vectors $$\lambda\vec{a} + 2\vec{b}$$ and $$3\vec{a} - \lambda\vec{b}$$ for $$\lambda\in[-1,3]$$; we seek the number of values of $$\lambda$$ for which these vectors are perpendicular.

Since perpendicularity means their dot product vanishes, we write

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

Expanding the dot product gives

$$3\lambda(\vec{a}\cdot\vec{a}) - \lambda^2(\vec{a}\cdot\vec{b}) + 6(\vec{b}\cdot\vec{a}) - 2\lambda(\vec{b}\cdot\vec{b}) = 0\,. $$

Because $$|\vec{a}|=|\vec{b}|=1$$ we have $$\vec{a}\cdot\vec{a}=1$$ and $$\vec{b}\cdot\vec{b}=1$$, while $$\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\frac{\pi}{3}=\frac12\,. $$

Substituting these results into the expanded dot product yields

$$3\lambda - \frac{\lambda^2}{2} + 6\cdot\frac12 - 2\lambda = 0\,, $$

which simplifies to

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

Multiplying both sides by $$-2$$ gives the quadratic equation

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

Solving this quadratic leads to

$$\lambda = \frac{2\pm\sqrt{4+24}}{2} = 1\pm\sqrt{7}\,. $$

Since $$\sqrt{7}\approx2.646$$, the two solutions are approximately $$1+\sqrt{7}\approx3.646$$ and $$1-\sqrt{7}\approx-1.646$$, neither of which lies in the interval $$[-1,3]$$.

Therefore there are no values of $$\lambda$$ in $$[-1,3]$$ that make the given vectors perpendicular, and hence the number of such values is 0.

Final Answer: 0.