If the plane `x- 3y +5z= d` passes through the point `(1, 2, 4),` then the intercepts cut by it on the axes of `x, y, z` are respectively-

Options

A. 15, -5, 3
B. 1, -5, 3
C. `-15, 5, -3`
D. `1, -6, 20`

Answer

To find the intercepts cut by the plane \( x - 3y + 5z = d \) on the axes of \( x, y, z \), we will follow these steps:

### Step 1: Determine the value of \( d \)
Since the plane passes through the point \( (1, 2, 4) \), we can substitute these coordinates into the plane equation to find \( d \).

\[ 1 - 3(2) + 5(4) = d \]

Calculating this gives:

\[ 1 - 6 + 20 = d \]
\[ d = 15 \]

### Step 2: Write the equation of the plane
Now that we have \( d \), we can rewrite the equation of the plane:

\[ x - 3y + 5z = 15 \]

### Step 3: Convert the plane equation to intercept form
The intercept form of a plane is given by:

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

To convert our plane equation into this form, we divide the entire equation by 15:

\[ \frac{x}{15} - \frac{3y}{15} + \frac{5z}{15} = 1 \]

This simplifies to:

\[ \frac{x}{15} - \frac{y}{5} + \frac{z}{3} = 1 \]

### Step 4: Identify the intercepts
From the equation in intercept form, we can identify the intercepts on the axes:

- The \( x \)-intercept \( (a, 0, 0) \) occurs when \( y = 0 \) and \( z = 0 \):
\[ x = 15 \Rightarrow (15, 0, 0) \]

- The \( y \)-intercept \( (0, b, 0) \) occurs when \( x = 0 \) and \( z = 0 \):
\[ -\frac{y}{5} = 1 \Rightarrow y = -5 \Rightarrow (0, -5, 0) \]

- The \( z \)-intercept \( (0, 0, c) \) occurs when \( x = 0 \) and \( y = 0 \):
\[ \frac{z}{3} = 1 \Rightarrow z = 3 \Rightarrow (0, 0, 3) \]

### Final Result
Thus, the intercepts cut by the plane on the axes of \( x, y, z \) are:

- \( x \)-intercept: \( (15, 0, 0) \)
- \( y \)-intercept: \( (0, -5, 0) \)
- \( z \)-intercept: \( (0, 0, 3) \)