The intercepts of the plane 5x - 3y + 6z = 60 | vedclass

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(C) To find the intercepts of the plane on the coordinate axes,we convert the equation into the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c

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$(12, -20, 10)$

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(C) To find the intercepts of the plane on the coordinate axes,we convert the equation into the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Given equation: $5x - 3y + 6z = 60$.Divide the entire equation by $60$:$\frac{5x}{60} - \frac{3y}{60} + \frac{6z}{60} = \frac{60}{60}$Simplifying the fractions:$\frac{x}{12} - \frac{y}{20} + \frac{z}{10} = 1$Rewriting in the standard intercept form:$\frac{x}{12} + \frac{y}{-20} + \frac{z}{10} = 1$Comparing this with $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,we get the intercepts as $a = 12$,$b = -20$,and $c = 10$.Thus,the intercepts are $(12, -20, 10)$.

The intercepts of the plane $5x - 3y + 6z = 60$ on the coordinate axes are

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