The distance of point `(1,-2,-3)` from plane `x-y+z=5` measured parallel to the line `x/2=y/3=z/(-6)` is

Options

A. 7
B. `1/7`
C. 1
D. 5

Answer

To find the distance of the point \( (1, -2, -3) \) from the plane \( x - y + z = 5 \) measured parallel to the line given by the equations \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \), we can follow these steps:

### Step 1: Write the equation of the line in parametric form
The line can be expressed in parametric form as:
\[ x = 2t, \quad y = 3t, \quad z = -6t \]
where \( t \) is a parameter.

### Step 2: Find a point on the line passing through the given point
We can express a point \( B \) on the line that passes through the point \( A(1, -2, -3) \) as:
\[ B = (1 + 2t, -2 + 3t, -3 - 6t) \]

### Step 3: Substitute the coordinates of point \( B \) into the plane equation
Substituting the coordinates of point \( B \) into the plane equation \( x - y + z = 5 \):
\[ (1 + 2t) - (-2 + 3t) + (-3 - 6t) = 5 \]
This simplifies to:
\[ 1 + 2t + 2 - 3t - 3 - 6t = 5 \]
Combining like terms gives:
\[ -7t = 5 - 1 - 2 + 3 \]
\[ -7t = 5 - 0 \]
\[ -7t = 5 \implies t = -\frac{5}{7} \]

### Step 4: Find the coordinates of point \( B \)
Now substituting \( t = -\frac{5}{7} \) back into the parametric equations for \( B \):
\[ x_B = 1 + 2\left(-\frac{5}{7}\right) = 1 - \frac{10}{7} = \frac{-3}{7} \]
\[ y_B = -2 + 3\left(-\frac{5}{7}\right) = -2 - \frac{15}{7} = \frac{-29}{7} \]
\[ z_B = -3 - 6\left(-\frac{5}{7}\right) = -3 + \frac{30}{7} = \frac{9}{7} \]
Thus, the coordinates of point \( B \) are \( \left( -\frac{3}{7}, -\frac{29}{7}, \frac{9}{7} \right) \).

### Step 5: Calculate the distance between points \( A \) and \( B \)
The distance \( d \) between points \( A(1, -2, -3) \) and \( B\left(-\frac{3}{7}, -\frac{29}{7}, \frac{9}{7}\right) \) is given by:
\[ d = \sqrt{\left(1 - \left(-\frac{3}{7}\right)\right)^2 + \left(-2 - \left(-\frac{29}{7}\right)\right)^2 + \left(-3 - \frac{9}{7}\right)^2} \]
Calculating each component:
1. \( 1 - \left(-\frac{3}{7}\right) = 1 + \frac{3}{7} = \frac{10}{7} \)
2. \( -2 - \left(-\frac{29}{7}\right) = -2 + \frac{29}{7} = \frac{15}{7} \)
3. \( -3 - \frac{9}{7} = -\frac{21}{7} - \frac{9}{7} = -\frac{30}{7} \)

Now substituting these into the distance formula:
\[ d = \sqrt{\left(\frac{10}{7}\right)^2 + \left(\frac{15}{7}\right)^2 + \left(-\frac{30}{7}\right)^2} \]
Calculating:
\[ d = \sqrt{\frac{100}{49} + \frac{225}{49} + \frac{900}{49}} = \sqrt{\frac{1225}{49}} = \frac{35}{7} = 5 \]

### Final Answer
The distance of the point \( (1, -2, -3) \) from the plane \( x - y + z = 5 \) measured parallel to the line \( \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} \) is \( 5 \).

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