The shortest distance of a point `(1, 2, -3)` from a plane making intercepts 1, 2 and 3 units on position X, Y and Z-axes respectively, is

Options

A. `2`
B. `0`
C. `(13)/(12)`
D. `(12)/(7)`

Answer

To find the shortest distance of the point \( (1, 2, -3) \) from the plane making intercepts 1, 2, and 3 units on the x, y, and z-axes respectively, we can follow these steps:

### Step 1: Write the equation of the plane
The intercept form of the equation of a plane is given by:

\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]

where \( a, b, c \) are the x, y, and z intercepts of the plane. Given that the intercepts are 1, 2, and 3, we can substitute these values into the equation:

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

### Step 2: Multiply through by the least common multiple (LCM)
To eliminate the denominators, multiply the entire equation by the LCM of the denominators (which is 6):

\[ 6 \left( \frac{x}{1} + \frac{y}{2} + \frac{z}{3} \right) = 6 \]

This simplifies to:

\[ 6x + 3y + 2z = 6 \]

### Step 3: Rearrange the equation into standard form
Rearranging the equation gives us:

\[ 6x + 3y + 2z - 6 = 0 \]

This is the equation of the plane.

### Step 4: Use the distance formula
The formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to a plane given by \( Ax + By + Cz + D = 0 \) is:

\[ D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

In our case, \( A = 6, B = 3, C = 2, D = -6 \), and the point is \( (1, 2, -3) \).

### Step 5: Substitute the values into the distance formula
Substituting the values into the formula gives:

\[ D = \frac{|6(1) + 3(2) + 2(-3) - 6|}{\sqrt{6^2 + 3^2 + 2^2}} \]

Calculating the numerator:

\[ D = \frac{|6 + 6 - 6 - 6|}{\sqrt{36 + 9 + 4}} \]

This simplifies to:

\[ D = \frac{|0|}{\sqrt{49}} = \frac{0}{7} = 0 \]

### Conclusion
The shortest distance of the point \( (1, 2, -3) \) from the plane is \( 0 \) units, which means the point lies on the plane.

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