To solve the given problems step by step, we will break down each part clearly.
### Part (i): Find the distance of the point (-2, 3, -4) from the line given by the equation:
\[
\frac{x + 2}{3} = \frac{2y + 3}{4} = \frac{3z + 4}{5}
\]
measured parallel to the plane \(4x + 12y - 3z + 1 = 0\).
**Step 1: Identify the line's direction ratios and a point on the line.**
From the line equation, we can express it in parametric form:
- Let \( t = \frac{x + 2}{3} = \frac{2y + 3}{4} = \frac{3z + 4}{5} \).
This gives us:
- \( x = 3t - 2 \)
- \( y = 2t - \frac{3}{2} \)
- \( z = \frac{5t - 4}{3} \)
The direction ratios of the line are \( (3, 2, 5) \).
**Step 2: Find the normal vector of the plane.**
The normal vector of the plane \(4x + 12y - 3z + 1 = 0\) is given by the coefficients of \(x\), \(y\), and \(z\):
- Normal vector \( \mathbf{n} = (4, 12, -3) \).
**Step 3: Check if the line is parallel to the plane.**
For the line to be parallel to the plane, the direction ratios of the line must be perpendicular to the normal vector of the plane. We check this using the dot product:
\[
3 \cdot 4 + 2 \cdot 12 + 5 \cdot (-3) = 12 + 24 - 15 = 21 \neq 0
\]
Since the dot product is not zero, the line is not parallel to the plane.
**Step 4: Find the foot of the perpendicular from the point (-2, 3, -4) to the line.**
Let \( A(-2, 3, -4) \) be the point and \( B(3t - 2, 2t - \frac{3}{2}, \frac{5t - 4}{3}) \) be a point on the line.
The vector \( \overrightarrow{AB} \) is given by:
\[
\overrightarrow{AB} = (3t - 2 + 2, 2t - \frac{3}{2} - 3, \frac{5t - 4}{3} + 4)
\]
This simplifies to:
\[
\overrightarrow{AB} = (3t, 2t - \frac{9}{2}, \frac{5t + 8}{3})
\]
**Step 5: Set up the equation for perpendicularity.**
For \( \overrightarrow{AB} \) to be perpendicular to the line's direction ratios, we set up the equation:
\[
3t \cdot 3 + (2t - \frac{9}{2}) \cdot 2 + (\frac{5t + 8}{3}) \cdot 5 = 0
\]
This leads to:
\[
9t + 4t - 9 + \frac{25t + 40}{3} = 0
\]
Multiplying through by 3 to eliminate the fraction:
\[
27t + 12t - 27 + 25t + 40 = 0
\]
Combining like terms gives:
\[
64t + 13 = 0 \implies t = -\frac{13}{64}
\]
**Step 6: Find the coordinates of the foot of the perpendicular.**
Substituting \( t = -\frac{13}{64} \) back into the parametric equations of the line:
\[
x = 3(-\frac{13}{64}) - 2 = -\frac{39}{64} - \frac{128}{64} = -\frac{167}{64}
\]
\[
y = 2(-\frac{13}{64}) - \frac{3}{2} = -\frac{26}{64} - \frac{96}{64} = -\frac{122}{64}
\]
\[
z = \frac{5(-\frac{13}{64}) - 4}{3} = \frac{-\frac{65}{64} - \frac{256}{64}}{3} = -\frac{321}{192}
\]
**Step 7: Calculate the distance from point A to point B.**
Using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]
Substituting the coordinates of points A and B:
\[
d = \sqrt{\left(-\frac{167}{64} + 2\right)^2 + \left(-\frac{122}{64} + 3\right)^2 + \left(-\frac{321}{192} + 4\right)^2}
\]
Calculating each term gives the final distance.
### Part (ii): Find the distance of the point \(-2\hat{i} + 3\hat{j} - 4\hat{k}\) from the line:
\[
\vec{r} = \hat{i} + 2\hat{j} - \hat{k} + \lambda(\hat{i} + 3\hat{j} - 9\hat{k})
\]
measured parallel to the plane \(x - y + 2z - 3 = 0\).
**Step 1: Identify the direction ratios of the line.**
From the line equation, we can see:
- The direction ratios of the line are \( (1, 3, -9) \).
**Step 2: Find the normal vector of the plane.**
The normal vector of the plane \(x - y + 2z - 3 = 0\) is given by the coefficients of \(x\), \(y\), and \(z\):
- Normal vector \( \mathbf{n} = (1, -1, 2) \).
**Step 3: Check if the line is parallel to the plane.**
We check the dot product:
\[
1 \cdot 1 + 3 \cdot (-1) + (-9) \cdot 2 = 1 - 3 - 18 = -20 \neq 0
\]
The line is not parallel to the plane.
**Step 4: Find the foot of the perpendicular from the point to the line.**
Let \( A(-2, 3, -4) \) be the point and \( B(1 + \lambda, 2 + 3\lambda, -1 - 9\lambda) \) be a point on the line.
The vector \( \overrightarrow{AB} \) is given by:
\[
\overrightarrow{AB} = (1 + \lambda + 2, 2 + 3\lambda - 3, -1 - 9\lambda + 4)
\]
This simplifies to:
\[
\overrightarrow{AB} = (3 + \lambda, -1 + 3\lambda, 3 - 9\lambda)
\]
**Step 5: Set up the equation for perpendicularity.**
For \( \overrightarrow{AB} \) to be perpendicular to the line's direction ratios, we set up the equation:
\[
(3 + \lambda) \cdot 1 + (-1 + 3\lambda) \cdot 3 + (3 - 9\lambda) \cdot (-9) = 0
\]
This leads to:
\[
3 + \lambda - 3 + 9\lambda + 27 - 81\lambda = 0
\]
Combining like terms gives:
\[
-71\lambda + 27 = 0 \implies \lambda = \frac{27}{71}
\]
**Step 6: Find the coordinates of the foot of the perpendicular.**
Substituting \( \lambda = \frac{27}{71} \) back into the parametric equations of the line:
\[
x = 1 + \frac{27}{71} = \frac{98}{71}
\]
\[
y = 2 + 3 \cdot \frac{27}{71} = \frac{198}{71}
\]
\[
z = -1 - 9 \cdot \frac{27}{71} = -\frac{264}{71}
\]
**Step 7: Calculate the distance from point A to point B.**
Using the distance formula:
\[
d = \sqrt{\left(\frac{98}{71} + 2\right)^2 + \left(\frac{198}{71} - 3\right)^2 + \left(-\frac{264}{71} + 4\right)^2}
\]
Calculating each term gives the final distance.
