Step 1: Understanding the Concept:
The given equation is of a plane in intercept form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
The intercepts on the x, y, and z axes are $a, b$, and $c$ respectively.
The vertices of the triangle are the points where the plane intersects the axes.
The area of this triangle can be found using vectors or by relating it to the areas of its projections on the coordinate planes.
Step 2: Key Formula or Approach:
If a plane has intercepts $a, b, c$ on the coordinate axes, the area $\Delta$ of the triangle formed by these intercept points is given by:
\[ \Delta = \frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2} \]
This comes from the relation $\Delta^2 = \Delta_{xy}^2 + \Delta_{yz}^2 + \Delta_{zx}^2$, where $\Delta_{xy} = \frac{1}{2}|ab|$ etc. are areas of projections.
Step 3: Detailed Explanation:
The equation of the plane is $\frac{x}{3} + \frac{y}{2} + \frac{z}{-4} = 1$.
Comparing with intercept form, we have:
x-intercept $a = 3 \Rightarrow$ point A is $(3, 0, 0)$
y-intercept $b = 2 \Rightarrow$ point B is $(0, 2, 0)$
z-intercept $c = -4 \Rightarrow$ point C is $(0, 0, -4)$
Using the formula for the area of the triangle:
\[ \Delta = \frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2} \]
Calculate the pairwise products:
$ab = 3 \cdot 2 = 6$
$bc = 2 \cdot (-4) = -8$
$ca = (-4) \cdot 3 = -12$
Now substitute these into the formula:
\[ \Delta = \frac{1}{2} \sqrt{(6)^2 + (-8)^2 + (-12)^2} \]
\[ \Delta = \frac{1}{2} \sqrt{36 + 64 + 144} \]
\[ \Delta = \frac{1}{2} \sqrt{244} \]
Simplify the radical:
\[ \sqrt{244} = \sqrt{4 \times 61} = 2\sqrt{61} \]
So, the area is:
\[ \Delta = \frac{1}{2} (2\sqrt{61}) = \sqrt{61} \]
Step 4: Final Answer:
The area of the triangle is $\sqrt{61}$ sq. units.