The equation of the plane whose intercepts on the coordinate axes are -4, 2 and 3, is
Answer & explanation
Correct answer: option 2
We know that the equation of a plane whose intercepts on the coordinate axes are a, b and c respectively, is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Here, a = -4, b = 2, and c = 3
So, the equation of the required plane is
$\frac{x}{-4}+\frac{y}{2}+\frac{z}{3}= 1 $ or , $ -3x + 6y + 4 z = 12 $