Find the values of lambda and mu for which the lines r = 3i - j + k + lambda(i + 2j - k) and r = 4i + 4j + k + mu(-i + j + 2k) intersect.

$$3\mathbf{i} - \mathbf{j} + \mathbf{k} + \lambda(\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 4\mathbf{i} + 4\mathbf{j} + \mathbf{k} + \mu(-\mathbf{i} + \mathbf{j} + 2\mathbf{k})$$

Expand and group terms by $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$:

$$(3 + \lambda)\mathbf{i} + (-1 + 2\lambda)\mathbf{j} + (1 - \lambda)\mathbf{k} = (4 - \mu)\mathbf{i} + (4 + \mu)\mathbf{j} + (1 + 2\mu)\mathbf{k}$$

For $\mathbf{i}$:

$$3 + \lambda = 4 - \mu$$ $$\lambda + \mu = 1 \quad \text{(Equation 1)}$$

For $\mathbf{j}$:

$$-1 + 2\lambda = 4 + \mu$$ $$2\lambda - \mu = 5 \quad \text{(Equation 2)}$$

For $\mathbf{k}$:

$$1 - \lambda = 1 + 2\mu$$ $$-\lambda - 2\mu = 0$$ $$\lambda + 2\mu = 0 \quad \text{(Equation 3)}$$

We can use Equation 1 and Equation 2.
Add Equation 1 and Equation 2:

$$(\lambda + \mu) + (2\lambda - \mu) = 1 + 5$$ $$3\lambda = 6$$ $$\lambda = 2$$

Substitute $\lambda = 2$ into Equation 1:

$$2 + \mu = 1$$ $$\mu = 1 - 2$$ $$\mu = -1$$

Substitute $\lambda = 2$ and $\mu = -1$ into Equation 3:

$$\lambda + 2\mu = 0$$ $$2 + 2(-1) = 0$$ $$2 - 2 = 0$$ $$0 = 0$$

The values satisfy all three equations, confirming that the lines intersect at these values.

The values are $\lambda = 2$ and $\mu = -1$.

Comparing this with the given options:
A $\lambda = 2$ and $\mu = 1$
B $\lambda = 2$ and $\mu = -1$
C $\lambda = -2$ and $\mu = -1$
D $\lambda = -2$ and $\mu = 1$

The correct option is B.

The final answer is $\boxed{\text{B}}$.