To find the distance of the point with position vector \(\mathbf{a} = -\hat{i} + 2\hat{j} + 6\hat{k}\) from the straight line passing through the point \((2, 3, -4)\) and parallel to the vector \(\mathbf{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}\), we can follow these steps:
### Step 1: Identify the given vectors
The point \(A\) has the position vector:
\[
\mathbf{A} = -\hat{i} + 2\hat{j} + 6\hat{k} \quad \text{(or in coordinates, } (-1, 2, 6)\text{)}
\]
The line passes through the point \(B\) with coordinates \((2, 3, -4)\) and is parallel to the vector \(\mathbf{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}\).
### Step 2: Write the parametric equations of the line
The parametric equations of the line can be expressed as:
\[
\begin{align*}
x &= 2 + 6t \\
y &= 3 + 3t \\
z &= -4 - 4t
\end{align*}
\]
where \(t\) is a parameter.
### Step 3: Find a point on the line
Let’s denote a point \(C\) on the line corresponding to parameter \(t\):
\[
C(t) = (2 + 6t, 3 + 3t, -4 - 4t)
\]
### Step 4: Find the vector \(\overrightarrow{AC}\)
The vector from point \(A\) to point \(C\) is given by:
\[
\overrightarrow{AC} = C(t) - A = \left( (2 + 6t) - (-1), (3 + 3t) - 2, (-4 - 4t) - 6 \right)
\]
Simplifying this gives:
\[
\overrightarrow{AC} = (3 + 6t, 1 + 3t, -10 - 4t)
\]
### Step 5: Find the direction vector of the line
The direction vector of the line is:
\[
\mathbf{b} = (6, 3, -4)
\]
### Step 6: Use the dot product to find the perpendicular distance
For the distance from point \(A\) to the line, we need to ensure that \(\overrightarrow{AC}\) is perpendicular to the direction vector \(\mathbf{b}\). This means:
\[
\overrightarrow{AC} \cdot \mathbf{b} = 0
\]
Calculating the dot product:
\[
(3 + 6t) \cdot 6 + (1 + 3t) \cdot 3 + (-10 - 4t) \cdot (-4) = 0
\]
Expanding this:
\[
18 + 36t + 3 + 9t + 40 + 16t = 0
\]
Combining like terms:
\[
(36t + 9t + 16t) + (18 + 3 + 40) = 0
\]
\[
61t + 61 = 0
\]
Solving for \(t\):
\[
t = -1
\]
### Step 7: Find the coordinates of point \(C\)
Substituting \(t = -1\) into the parametric equations:
\[
C(-1) = (2 + 6(-1), 3 + 3(-1), -4 - 4(-1)) = (-4, 0, 0)
\]
### Step 8: Calculate the distance \(d\) from point \(A\) to point \(C\)
The distance \(d\) is given by:
\[
d = \sqrt{((-4) - (-1))^2 + (0 - 2)^2 + (0 - 6)^2}
\]
Calculating each term:
\[
d = \sqrt{(-3)^2 + (-2)^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7
\]
### Final Answer
The distance of the point from the line is:
\[
\boxed{7}
\]
