The sum of Y and Z intercepts of the plane $3 x + 4 y - 6 z = 12$ is ___________.

10

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4

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1

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5

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Solution

The correct option is B $1$
$3 x + 4 y - 6 z = 12$
$∴ \frac{3 x}{12} + \frac{4 y}{12} + \frac{(- 6) z}{12} = 1$
$∴ \frac{x}{4} + \frac{y}{3} + \frac{z}{(- 2)} = 1$
$∴$ y intercept $b = 3$ and z intercept $c = - 2$
$∴$ y intercept $+$ z intercept $= 3 + (- 2) = 1$

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