The distance between the line $$\vec{r} = 2\hat{i} - 2\hat{j} + 3\hat{k} + \lambda(\hat{i} - \hat{j} + 4\hat{k})$$ and the plane $$\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$$ is

The line is $$\vec r = 2\hat i - 2\hat j + 3\hat k + \lambda\,(\,\hat i - \hat j + 4\hat k\,)$$, so

direction vector of the line: $$\vec d = \hat i - \hat j + 4\hat k$$

The plane is $$\vec r \,\cdot\, (\hat i + 5\hat j + \hat k) = 5$$, so

normal vector of the plane: $$\vec n = \hat i + 5\hat j + \hat k$$

First check whether the line is parallel to the plane. Compute $$\vec d \cdot \vec n$$:

$$\vec d \cdot \vec n = 1 \times 1 + (-1)\times 5 + 4 \times 1 = 1 - 5 + 4 = 0$$

Because the dot product is zero, $$\vec d \perp \vec n$$; therefore the line is parallel to the plane. Hence every point on the line is at the same perpendicular distance from the plane. We can use any convenient point on the line.

Take the point corresponding to $$\lambda = 0$$:

$$P(2,\,-2,\,3)$$

The Cartesian form of the plane is

$$x + 5y + z - 5 = 0 \quad -(1)$$

The perpendicular distance $$D$$ from a point $$(x_0, y_0, z_0)$$ to the plane $$Ax + By + Cz + D_0 = 0$$ is

$$D = \frac{\lvert Ax_0 + By_0 + Cz_0 + D_0 \rvert}{\sqrt{A^2 + B^2 + C^2}}$$

Here $$A = 1,\; B = 5,\; C = 1,\; D_0 = -5$$ and $$(x_0, y_0, z_0) = (2, -2, 3)$$.

Numerator:

$$\lvert 1(2) + 5(-2) + 1(3) - 5 \rvert = \lvert 2 - 10 + 3 - 5 \rvert = \lvert -10 \rvert = 10$$

Denominator:

$$\sqrt{1^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3}$$

Hence

$$D = \frac{10}{3\sqrt{3}}$$

Option B which is: $$\frac{10}{3\sqrt{3}}$$