1. Since the vectors are mutually orthogonal, each pair must have dot product zero:

$\vec a \cdot \vec b = 0, \quad \vec a \cdot \vec c = 0, \quad \vec b \cdot \vec c = 0$

Given:

$\vec a = \hat i - \hat j + 2\hat k = (1,-1,2)$ $\vec b = 2\hat i + 4\hat j + \hat k = (2,4,1)$ $\vec c = \lambda \hat i + \hat j + \mu \hat k = (\lambda,1,\mu)$

2. First check that $\vec a$ and $\vec b$ are orthogonal:

$\vec a \cdot \vec b = (1)(2) + (-1)(4) + (2)(1) = 2 - 4 + 2 = 0$

So this condition is satisfied.

3. Now use $\vec a \cdot \vec c = 0$:

$(1)(\lambda) + (-1)(1) + (2)(\mu) = 0$ $\lambda - 1 + 2\mu = 0$ $\lambda + 2\mu = 1 \quad \text{...(1)}$

4. Use $\vec b \cdot \vec c = 0$:

$(2)(\lambda) + (4)(1) + (1)(\mu) = 0$ $2\lambda + 4 + \mu = 0$ $2\lambda + \mu = -4 \quad \text{...(2)}$

5. Solve the system:

From (1):

$\lambda = 1 - 2\mu$

Substitute into (2):

$2(1 - 2\mu) + \mu = -4$ $2 - 4\mu + \mu = -4$ $2 - 3\mu = -4$ $-3\mu = -6$ $\mu = 2$

Then

$\lambda = 1 - 2(2) = 1 - 4 = -3$

So,

$(\lambda, \mu) = (-3, 2)$

6. Compare with options:

This matches Option D.