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in Three Dimensional Geometry by (4.0k points)
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(i) Find the equation of the Plane with intercepts 2, 3, 4 on X, Y, Z axes respectively.) 

(ii) Find the distance of the point (- 1, – 2,3) from the Plane \(\bar{r}\) (2i – 3 j + 4k) = 4

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Best answer

(i) The equation of the plane is

\(\frac{x}{2}+\frac{y}{3}+\frac{z}{4} = 1\)

(ii) The equation of the plane is 2x – 3y + 4z = 4

Hence the distance = \(\frac{2(-1)-3(-2)+4(3)-4}{\sqrt{4+9+16}}\)

\( = \frac{-2+6+12-4}{\sqrt{4+9+16}}\)

\(\frac{12}{\sqrt{29}}\)

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