Find the equation of the plane with intercepts and on and axes respectively.
Find the equation of the plane with intercepts $3, -4$ and $2$ on $x, y$ and $z$ axes respectively.
Answer
Answer
AIWritten by AI (gemini-2.5-pro) - it can contain mistakes.
The equation of a plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
With intercepts $a = 3$, $b = -4$, $c = 2$:
$\frac{x}{3} - \frac{y}{4} + \frac{z}{2} = 1$.
Multiplying by 12 gives: $4x - 3y + 6z = 12$ (or $4x - 3y + 6z - 12 = 0$).
From ISC 2026 Mathematics Paper 1, question 15(v).