The distance of the point $$-\hat{i} + 2\hat{j} + 6\hat{k}$$ from the straight line that passes through the point $$2\hat{i} + 3\hat{j} - 4\hat{k}$$ and is parallel to the vector $$6\hat{i} + 3\hat{j} - 4\hat{k}$$ is

Let the given point be $$P(-1,\,2,\,6)$$.

The required line passes through $$A(2,\,3,\,-4)$$ and is parallel to the vector $$\vec{b}=6\hat{i}+3\hat{j}-4\hat{k}\; \Rightarrow\; \vec{b}=(6,\,3,\,-4)$$.

For the distance of a point from a line in 3-D, we use
$$\text{Distance} \; d=\dfrac{\left|\; \overrightarrow{AP}\times\vec{b}\;\right|}{|\vec{b}|},$$
where $$\overrightarrow{AP}=P-A$$.

Compute $$\overrightarrow{AP}:$$
$$\overrightarrow{AP}=(-1-2,\;2-3,\;6-(-4))=(-3,\,-1,\,10).$$

Now find the cross product $$\overrightarrow{AP}\times\vec{b}:$$
$$\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\[2pt] -3&-1&10\\[2pt] 6&3&-4 \end{vmatrix} = \hat{i}\bigl((-1)(-4)-10\cdot3\bigr) -\hat{j}\bigl((-3)(-4)-10\cdot6\bigr) +\hat{k}\bigl((-3)(3)-(-1)\cdot6\bigr).$$
Evaluating each component:
$$\hat{i}(-26)+\hat{j}(48)+\hat{k}(-3)\; \Longrightarrow\; (-26,\,48,\,-3).$$

Magnitude of the cross product:
$$\left|(-26,\,48,\,-3)\right| =\sqrt{(-26)^2+48^2+(-3)^2} =\sqrt{676+2304+9} =\sqrt{2989}.$$

Magnitude of the direction vector $$\vec{b}:$$
$$|\vec{b}|=\sqrt{6^2+3^2+(-4)^2} =\sqrt{36+9+16} =\sqrt{61}.$$

Hence the required distance is
$$d=\frac{\sqrt{2989}}{\sqrt{61}} =\sqrt{\frac{2989}{61}}.$$

Notice that $$61\times49=2989,$$ so
$$d=\sqrt{49}=7.$$

Option C which is: 7