1. Given vectors

$\vec a=\hat i+\lambda \hat j-3\hat k=(1,\lambda,-3),\qquad \vec b=3\hat i-\hat j+2\hat k=(3,-1,2)$

We are told that $\vec a+\vec b$ and $\vec a-\vec b$ are perpendicular.

2. Use perpendicularity condition

If two vectors are perpendicular, then their dot product is zero:

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

Now,

$(\vec a+\vec b)\cdot(\vec a-\vec b)=\vec a\cdot\vec a-\vec b\cdot\vec b=|\vec a|^2-|\vec b|^2$

So,

$|\vec a|^2=|\vec b|^2$

Compute both:

$|\vec a|^2=1^2+\lambda^2+(-3)^2=1+\lambda^2+9=\lambda^2+10$

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

Hence,

$\lambda^2+10=14 \Rightarrow \lambda^2=4$

Given $\lambda>0$, we get

$\lambda=2$

3. Find $\cos\theta$

Angle $\theta$ is between $\vec a$ and $\vec b$, so

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

With $\lambda=2$,

$\vec a=(1,2,-3)$

Now compute dot product:

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

Also,

$|\vec a|=|\vec b|=\sqrt{14}$

Therefore,

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

4. Compute $(14\cos\theta)^2$

$14\cos\theta=14\left(\frac{-5}{14}\right)=-5$

Thus,

$(14\cos\theta)^2=(-5)^2=25$

5. Match with options

The correct option is:

$\boxed{25}$

So, Option A is correct.