Distance of the point `(1,-2,3)` from the plane `x-y+z = 5` measured parallel to the line whose direction cosines are proportional to `2,3,-6` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the distance of the point \( (1, -2, 3) \) from the plane \( x - y + z = 5 \) measured parallel to the line whose direction cosines are proportional to \( 2, 3, -6 \), we can follow these steps:

### Step 1: Write the equation of the line
The direction ratios of the line are \( 2, 3, -6 \). The point on the line is given as \( (1, -2, 3) \). The parametric equations of the line can be expressed as:

\[ \frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{-6} = \lambda \]

From this, we can express \( x, y, z \) in terms of \( \lambda \):

\[ x = 2\lambda + 1 \]
\[ y = 3\lambda - 2 \]
\[ z = -6\lambda + 3 \]

### Step 2: Substitute into the plane equation
The equation of the plane is given by:

\[ x - y + z = 5 \]

Substituting the parametric equations into the plane equation:

\[ (2\lambda + 1) - (3\lambda - 2) + (-6\lambda + 3) = 5 \]

### Step 3: Simplify the equation
Now, simplify the equation:

\[ 2\lambda + 1 - 3\lambda + 2 - 6\lambda + 3 = 5 \]

Combine like terms:

\[ (2\lambda - 3\lambda - 6\lambda) + (1 + 2 + 3) = 5 \]
\[ -7\lambda + 6 = 5 \]

### Step 4: Solve for \( \lambda \)
Now, isolate \( \lambda \):

\[ -7\lambda = 5 - 6 \]
\[ -7\lambda = -1 \]
\[ \lambda = \frac{1}{7} \]

### Step 5: Find the coordinates of the intersection point
Now substitute \( \lambda = \frac{1}{7} \) back into the parametric equations to find the coordinates of the intersection point:

\[ x = 2\left(\frac{1}{7}\right) + 1 = \frac{2}{7} + 1 = \frac{9}{7} \]
\[ y = 3\left(\frac{1}{7}\right) - 2 = \frac{3}{7} - 2 = \frac{3}{7} - \frac{14}{7} = -\frac{11}{7} \]
\[ z = -6\left(\frac{1}{7}\right) + 3 = -\frac{6}{7} + 3 = -\frac{6}{7} + \frac{21}{7} = \frac{15}{7} \]

So the intersection point \( Q \) is \( \left( \frac{9}{7}, -\frac{11}{7}, \frac{15}{7} \right) \).

### Step 6: Calculate the distance from point \( P \) to point \( Q \)
Now, we need to calculate the distance between the point \( P(1, -2, 3) \) and the intersection point \( Q \):

\[ \text{Distance} = \sqrt{(1 - \frac{9}{7})^2 + (-2 + \frac{11}{7})^2 + (3 - \frac{15}{7})^2} \]

Calculating each term:

1. \( 1 - \frac{9}{7} = \frac{7}{7} - \frac{9}{7} = -\frac{2}{7} \)
2. \( -2 + \frac{11}{7} = -\frac{14}{7} + \frac{11}{7} = -\frac{3}{7} \)
3. \( 3 - \frac{15}{7} = \frac{21}{7} - \frac{15}{7} = \frac{6}{7} \)

Now substituting back into the distance formula:

\[ \text{Distance} = \sqrt{\left(-\frac{2}{7}\right)^2 + \left(-\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2} \]
\[ = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1 \]

### Final Answer
The distance of the point \( (1, -2, 3) \) from the plane \( x - y + z = 5 \) measured parallel to the line whose direction cosines are proportional to \( 2, 3, -6 \) is \( 1 \).

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