We are given the vectors

$ \vec{a} = \hat{i} + \lambda\,\hat{j} - 3\,\hat{k} \quad \text{and} \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}, $

with $\lambda > 0$, and we are told that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular. Our goal is to find $(14 \cos\theta)^2$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Step 1. Use the Perpendicularity Condition

Two vectors are perpendicular if their dot product is zero. So, we require:

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

First, compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$:

$ \vec{a}+\vec{b} = (\hat{i}+3\,\hat{i})+(\lambda\,\hat{j}-\hat{j})+(-3\,\hat{k}+2\,\hat{k}) = 4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}. $

$ \vec{a}-\vec{b} = (\hat{i}-3\,\hat{i})+(\lambda\,\hat{j}+ \hat{j})+(-3\,\hat{k}-2\,\hat{k}) = -2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}. $

Now, take their dot product:

$ (4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}) \cdot (-2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}) $

$ = 4(-2) + (\lambda-1)(\lambda+1) + (-1)(-5) = -8 + (\lambda^2 - 1) + 5. $

$ = \lambda^2 - 4. $

Setting the dot product to zero:

$ \lambda^2 - 4 = 0 \quad \Longrightarrow \quad \lambda^2 = 4. $

Since $\lambda > 0$, we have:

$ \lambda = 2. $

Step 2. Find the Angle Between $\vec{a}$ and $\vec{b}$

Now that we know $\lambda = 2$, the vectors become:

$ \vec{a} = \hat{i} + 2\,\hat{j} - 3\,\hat{k}, \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}. $

The cosine of the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is given by:

$ \cos\theta = \frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\| \|\vec{b}\|}. $

Calculate $\vec{a} \cdot \vec{b}$:

$ \vec{a}\cdot\vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5. $

Calculate the magnitudes:

$ \|\vec{a}\| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}. $

$ \|\vec{b}\| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}. $

Thus,

$ \cos\theta = \frac{-5}{\sqrt{14}\cdot\sqrt{14}} = \frac{-5}{14}. $

Step 3. Compute $(14\cos\theta)^2$

$ (14\cos\theta)^2 = \left(14\cdot\frac{-5}{14}\right)^2 = (-5)^2 = 25. $

Final Answer

The value of $(14\cos\theta)^2$ is:

$ \boxed{25}. $