Algebraic sum of the intercepts made by the plane `x+3y-4z+6=0` on the axes is

Options

A. `-13//2`
B. `19//2`
C. `-22//3`
D. `26//3`

Answer

To find the algebraic sum of the intercepts made by the plane \( x + 3y - 4z + 6 = 0 \) on the axes, we will follow these steps:

### Step 1: Identify the Intercepts
The intercepts on the axes can be found by setting the other two variables to zero in the equation of the plane.

### Step 2: Calculate the X-Intercept
To find the x-intercept, set \( y = 0 \) and \( z = 0 \):
\[ x + 3(0) - 4(0) + 6 = 0 \implies x + 6 = 0 \implies x = -6 \]
Thus, the x-intercept is \( -6 \).

### Step 3: Calculate the Y-Intercept
To find the y-intercept, set \( x = 0 \) and \( z = 0 \):
\[ 0 + 3y - 4(0) + 6 = 0 \implies 3y + 6 = 0 \implies 3y = -6 \implies y = -2 \]
Thus, the y-intercept is \( -2 \).

### Step 4: Calculate the Z-Intercept
To find the z-intercept, set \( x = 0 \) and \( y = 0 \):
\[ 0 + 3(0) - 4z + 6 = 0 \implies -4z + 6 = 0 \implies -4z = -6 \implies z = \frac{6}{4} = \frac{3}{2} \]
Thus, the z-intercept is \( \frac{3}{2} \).

### Step 5: Sum the Intercepts
Now, we will find the algebraic sum of the intercepts:
\[ \text{Algebraic Sum} = x + y + z = -6 + (-2) + \frac{3}{2} \]
To add these, we first convert \( -6 \) and \( -2 \) into a common denominator with \( \frac{3}{2} \):
\[ -6 = -\frac{12}{2}, \quad -2 = -\frac{4}{2} \]
Now, we can sum them:
\[ -\frac{12}{2} - \frac{4}{2} + \frac{3}{2} = -\frac{12 + 4 - 3}{2} = -\frac{13}{2} \]

### Final Answer
The algebraic sum of the intercepts made by the plane on the axes is:
\[ \boxed{-\frac{13}{2}} \]