Find the vector equation of a line passing through the point with position vector `(2 hati - 3 hatj - 5 hatk)` and perpendicular to the plane `vec(r). (6 hati - 3 hatj - 5 hatk)` + 2 = 0. Also find the point of intersection of this line and the plane.

Answer

To solve the problem, we need to find the vector equation of a line that passes through a given point and is perpendicular to a specified plane. We will also find the point of intersection of this line and the plane.

### Step 1: Identify the given information
- The position vector of the point through which the line passes is \( \mathbf{a} = 2\hat{i} - 3\hat{j} - 5\hat{k} \).
- The equation of the plane is given as \( \mathbf{r} \cdot (6\hat{i} - 3\hat{j} - 5\hat{k}) + 2 = 0 \).

### Step 2: Determine the normal vector of the plane
The normal vector \( \mathbf{n} \) of the plane can be extracted from the equation of the plane. Here, \( \mathbf{n} = 6\hat{i} - 3\hat{j} - 5\hat{k} \).

### Step 3: Write the vector equation of the line
The vector equation of a line can be expressed as:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
where \( \mathbf{a} \) is the position vector of a point on the line, \( \lambda \) is a scalar parameter, and \( \mathbf{b} \) is the direction vector of the line. Since the line is perpendicular to the plane, the direction vector \( \mathbf{b} \) is the same as the normal vector \( \mathbf{n} \).

Thus, the vector equation of the line is:
\[ \mathbf{r} = (2\hat{i} - 3\hat{j} - 5\hat{k}) + \lambda (6\hat{i} - 3\hat{j} - 5\hat{k}) \]

### Step 4: Find the point of intersection of the line and the plane
To find the point of intersection, we need to substitute the parametric equations of the line into the equation of the plane.

From the line equation, we can express the coordinates \( x, y, z \) in terms of \( \lambda \):
- \( x = 2 + 6\lambda \)
- \( y = -3 - 3\lambda \)
- \( z = -5 - 5\lambda \)

Substituting these into the plane equation \( 6x - 3y - 5z + 2 = 0 \):
\[ 6(2 + 6\lambda) - 3(-3 - 3\lambda) - 5(-5 - 5\lambda) + 2 = 0 \]

Expanding this:
\[ 12 + 36\lambda + 9 + 9\lambda + 25 + 25\lambda + 2 = 0 \]
Combining like terms:
\[ (36\lambda + 9\lambda + 25\lambda) + (12 + 9 + 25 + 2) = 0 \]
\[ 70\lambda + 48 = 0 \]
Solving for \( \lambda \):
\[ 70\lambda = -48 \implies \lambda = -\frac{48}{70} = -\frac{24}{35} \]

### Step 5: Calculate the coordinates of the point of intersection
Substituting \( \lambda = -\frac{24}{35} \) back into the parametric equations:
- \( x = 2 + 6\left(-\frac{24}{35}\right) = 2 - \frac{144}{35} = \frac{70 - 144}{35} = -\frac{74}{35} \)
- \( y = -3 - 3\left(-\frac{24}{35}\right) = -3 + \frac{72}{35} = -\frac{105 - 72}{35} = -\frac{33}{35} \)
- \( z = -5 - 5\left(-\frac{24}{35}\right) = -5 + \frac{120}{35} = -\frac{175 - 120}{35} = -\frac{55}{35} \)

Thus, the point of intersection is:
\[ \left(-\frac{74}{35}, -\frac{33}{35}, -\frac{55}{35}\right) \]

### Final Result
The vector equation of the line is:
\[ \mathbf{r} = (2\hat{i} - 3\hat{j} - 5\hat{k}) + \lambda (6\hat{i} - 3\hat{j} - 5\hat{k}) \]
And the point of intersection of the line and the plane is:
\[ \left(-\frac{74}{35}, -\frac{33}{35}, -\frac{55}{35}\right) \]