The line of intersection of the planes $$\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$$ and $$\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$$, is:

We have two planes given by the vector equations

$$\vec r \cdot (3\hat i-\hat j+\hat k)=1 \qquad$$ and $$\qquad \vec r \cdot (\hat i+4\hat j-2\hat k)=2.$$

Writing the dot products in Cartesian form, they become

$$3x-y+z=1 \qquad\qquad (1)$$

$$x+4y-2z=2 \qquad\qquad (2).$$

Every line of intersection of two planes is parallel to the vector which is perpendicular to both normals of the planes. The normal of the first plane is $$\vec n_1 = 3\hat i-\hat j+\hat k$$ and the normal of the second plane is $$\vec n_2 = \hat i+4\hat j-2\hat k.$$

To get a vector parallel to the required line, we use the formula for the cross product of the normals:

$$\vec d = \vec n_1 \times \vec n_2.$$

By definition of the cross product,

$$ \vec d = \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -1 & 1\\ 1 & 4 & -2 \end{vmatrix}. $$

Expanding this determinant step by step, we obtain

$$ \vec d = \hat i\bigl((-1)(-2)-1\cdot4\bigr) - \hat j\bigl(3(-2)-1\cdot1\bigr) + \hat k\bigl(3\cdot4-(-1)\cdot1\bigr). $$

Simplifying each component carefully,

$$ \begin{aligned} \hat i &: (-1)(-2)-1\cdot4 = 2-4 = -2,\\ -\hat j &: 3(-2)-1\cdot1 = -6-1 = -7 \;\;\text{and the minus sign in front turns it into } +7,\\ \hat k &: 3\cdot4-(-1)\cdot1 = 12+1 = 13. \end{aligned} $$

So

$$\vec d = -2\hat i + 7\hat j + 13\hat k.$$

The vector $$\vec d$$ can be multiplied by $$-1$$ without changing the direction of the line, giving the equally valid direction vector

$$\vec d_1 = 2\hat i - 7\hat j - 13\hat k.$$

We next need a single point that lies on both planes. Let us search for such a point by solving equations (1) and (2) for $$x$$, $$y$$, and $$z$$. A common trick is to set one variable conveniently and solve for the other two. We choose

$$z = 0.$$

Substituting $$z=0$$ into (1) gives

$$3x - y = 1 \qquad\qquad (1').$$

Substituting $$z=0$$ into (2) gives

$$x + 4y = 2 \qquad\qquad (2').$$

From (1') we can express $$y$$ in terms of $$x$$:

$$y = 3x - 1.$$

Substituting this value of $$y$$ into (2'):

$$x + 4(3x-1) = 2.$$

Expanding and combining like terms,

$$x + 12x - 4 = 2,$$

$$13x - 4 = 2,$$

$$13x = 6,$$

$$x = \dfrac{6}{13}.$$

Now compute $$y$$:

$$y = 3\left(\dfrac{6}{13}\right)-1 = \dfrac{18}{13} - \dfrac{13}{13} = \dfrac{5}{13}.$$

Because we chose $$z=0$$, the required point is

$$P\Bigl(\dfrac{6}{13},\,\dfrac{5}{13},\,0\Bigr).$$

With point $$P$$ and direction vector $$\vec d_1 = 2\hat i - 7\hat j - 13\hat k$$, the vector equation of the line is

$$\vec r = \Bigl(\dfrac{6}{13}\hat i + \dfrac{5}{13}\hat j\Bigr) + t\,(2\hat i - 7\hat j - 13\hat k), \quad t\in\mathbb R.$$

Translating this into parametric Cartesian form,

$$x = \dfrac{6}{13} + 2t,\qquad y = \dfrac{5}{13} - 7t,\qquad z = -13t.$$

Eliminating the parameter $$t$$, we divide each coordinate’s displacement by its corresponding direction component, giving the symmetric form

$$\frac{x-\dfrac{6}{13}}{2} = \frac{y-\dfrac{5}{13}}{-7} = \frac{z}{-13}.$$

This expression is exactly the description given in Option C.

Hence, the correct answer is Option C.