The volume of the tetrahedron included between the plane `3x+4y-5z-60=0` and the co-odinate planes is

Options

A. `60`
B. `600`
C. `720`
D. `400`

Answer

To find the volume of the tetrahedron formed by the plane \(3x + 4y - 5z - 60 = 0\) and the coordinate planes, we can follow these steps:

### Step 1: Find the intercepts of the plane with the coordinate axes.
To find the intercepts, we set two of the variables to zero and solve for the third.

1. **X-intercept**: Set \(y = 0\) and \(z = 0\):
\[ 3x - 60 = 0 \implies x = 20 \]
So, the X-intercept is \((20, 0, 0)\).

2. **Y-intercept**: Set \(x = 0\) and \(z = 0\):
\[ 4y - 60 = 0 \implies y = 15 \]
So, the Y-intercept is \((0, 15, 0)\).

3. **Z-intercept**: Set \(x = 0\) and \(y = 0\):
\[ -5z - 60 = 0 \implies z = -12 \]
So, the Z-intercept is \((0, 0, -12)\).

### Step 2: Identify the vertices of the tetrahedron.
The vertices of the tetrahedron formed by the intercepts and the origin are:
- \(A(0, 0, 0)\)
- \(B(20, 0, 0)\)
- \(C(0, 15, 0)\)
- \(D(0, 0, -12)\)

### Step 3: Calculate the vectors.
Now, we can find the vectors from the origin \(A\) to the other points:
- Vector \( \vec{AB} = B - A = (20, 0, 0) - (0, 0, 0) = 20\hat{i} \)
- Vector \( \vec{AC} = C - A = (0, 15, 0) - (0, 0, 0) = 15\hat{j} \)
- Vector \( \vec{AD} = D - A = (0, 0, -12) - (0, 0, 0) = -12\hat{k} \)

### Step 4: Calculate the volume of the tetrahedron.
The volume \(V\) of the tetrahedron can be calculated using the formula:
\[ V = \frac{1}{6} | \vec{AB} \cdot (\vec{AC} \times \vec{AD}) | \]

#### Step 4.1: Calculate the cross product \( \vec{AC} \times \vec{AD} \).
\[ \vec{AC} \times \vec{AD} = (15\hat{j}) \times (-12\hat{k}) = -180\hat{i} \]

#### Step 4.2: Calculate the dot product \( \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \).
\[ \vec{AB} \cdot (-180\hat{i}) = 20\hat{i} \cdot (-180\hat{i}) = -3600 \]

#### Step 4.3: Calculate the volume.
\[ V = \frac{1}{6} | -3600 | = \frac{3600}{6} = 600 \]

### Final Answer:
The volume of the tetrahedron is \(600 \, \text{cubic units}\).

---