You are not asked to **derive** the equation of the plane, just to show that this is the equation of that plane.

Since \(\displaystyle \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= 1\) is linear in each variable, it represents a plane.

If x= a, y= 0, z= 0, then \(\displaystyle \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= \frac{a}{a}+ \frac{0}{b}+ \frac{0}{c}= 1\) so the plane contains (a, 0, 0).

If x= 0, y= b, z= 0, then \(\displaystyle \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= \frac{0}{a}+ \frac{b}{b}+ \frac{0}{c}= 1\) so the plane contains (0, b, 0).

If x= 0, y= 0, z= c, then \(\displaystyle \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= \frac{0}{a}+ \frac{0}{b}+ \frac{c}{c}= 1\) so the plane contains (0, 0, c).

Since a plane is determined by three points, this is the required plane.