If the vectors $$\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$$, $$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$$ and $$\vec{c} = \lambda \hat{i} + \hat{j} + \mu \hat{k}$$ are mutually orthogonal, then $$(\lambda, \mu) =$$

For three vectors to be mutually orthogonal, the dot product of every pair must vanish.

First verify that $$\vec a$$ and $$\vec b$$ are orthogonal:
$$\vec a \cdot \vec b = (1)(2) + (-1)(4) + (2)(1) = 2 - 4 + 2 = 0$$
Hence $$\vec a \perp \vec b$$ as required.

Let $$\vec c = \lambda \hat i + \hat j + \mu \hat k$$. Orthogonality of $$\vec c$$ with the other two vectors gives two linear equations.

1. $$\vec a \cdot \vec c = 0$$:
$$\bigl(\hat i - \hat j + 2\hat k\bigr) \cdot \bigl(\lambda \hat i + \hat j + \mu \hat k\bigr) = 0$$
$$\Rightarrow \lambda - 1 + 2\mu = 0$$ $$-(1)$$

2. $$\vec b \cdot \vec c = 0$$:
$$\bigl(2\hat i + 4\hat j + \hat k\bigr) \cdot \bigl(\lambda \hat i + \hat j + \mu \hat k\bigr) = 0$$
$$\Rightarrow 2\lambda + 4 + \mu = 0$$ $$-(2)$$

Solve the simultaneous equations (1) and (2). From (1): $$\lambda + 2\mu = 1$$ $$-(3)$$ From (2): $$2\lambda + \mu = -4$$ $$-(4)$$

Multiply (3) by 2 and subtract (4):
$$2\lambda + 4\mu = 2$$
$$\underline{\phantom{2\lambda + 4\mu = 2}}\;<-\;(2\lambda + \mu = -4)$$
$$3\mu = 6 \;\Rightarrow\; \mu = 2$$

Substitute $$\mu = 2$$ into (3):
$$\lambda + 2(2) = 1 \;\Rightarrow\; \lambda = -3$$

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

Option D which is: $$(-3, 2)$$