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If the plane $x-3 y+5 z=d$ passes through the point $(1,2,4)$, then the lengths of intercepts cut by it on the axes of $x, y, z$ are respectively
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$-15,5,-3$
If plane $x-3 y+5 z=d$ passes through point $(1,2,4)$. Then $1-6+20=d \Rightarrow d=15$
$\therefore$ Plane, $x-3 y+5 z=15 \Rightarrow \frac{x}{15}+\frac{y}{-5}+\frac{z}{3}=1$
Hence length of intercept cut by it on the axes $(x, y, z)$ are respectively $(15,-5,3)$.
$\therefore$ Plane, $x-3 y+5 z=15 \Rightarrow \frac{x}{15}+\frac{y}{-5}+\frac{z}{3}=1$
Hence length of intercept cut by it on the axes $(x, y, z)$ are respectively $(15,-5,3)$.
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