Question

The Cartesian equation of a plane which passes through the points $\text{A}(2, 2, 2)$ and making equal nonzero intercepts on the co-ordinate axes is

• A. $x + y + z = 6$
• B. $x - 2y + z = 0$
• C. $2x + y + z = 7$
• D. $x - y + z = 2$

Hint:

Use a quick point substitution check to save time! Plug the coordinates $(2,2,2)$ straight into the options:
For (A): $2 + 2 + 2 = 6$ (True)
For (B): $2 - 4 + 2 = 0$ (True, but does not have equal non-zero intercepts)
Checking options using point validation is an excellent way to confirm your answer!

Answer 1

The Correct Option is A

Step 1: Use the intercept form.
A plane making intercepts $a$, $b$, $c$ on the axes is $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$.
Step 2: Apply the equal-intercepts condition.
The intercepts are equal and non-zero, so set $a=b=c$. The equation becomes $\dfrac{x}{a}+\dfrac{y}{a}+\dfrac{z}{a}=1$.
Step 3: Clear the denominator.
Multiply through by $a$: $x+y+z=a$. So every such plane has the form $x+y+z=\text{constant}$.
Step 4: Use the given point.
The plane passes through $A(2,2,2)$, so this point must satisfy the equation.
Step 5: Solve for the constant.
Substitute: $2+2+2=a$, hence $a=6$.
Step 6: Write the final equation.
Therefore the plane is $x+y+z=6$, matching option (1).
\[ \boxed{x+y+z=6} \]