A plane E_{1} makes intercepts 1, -3, 4 on th | vedclass

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(C) The intercept form of the plane $E_{1}$ with intercepts $a=1, b=-3, c=4$ is $\frac{x}{1} + \frac{y}{-3} + \frac{z}{4} = 1$. Multiplying by $12$,we

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$\frac{x}{1}-\frac{y}{3}+\frac{z}{4}+2=0$

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(C) The intercept form of the plane $E_{1}$ with intercepts $a=1, b=-3, c=4$ is $\frac{x}{1} + \frac{y}{-3} + \frac{z}{4} = 1$. Multiplying by $12$,we get $12x - 4y + 3z = 12$,or $12x - 4y + 3z - 12 = 0$. The normal vector to this plane is $\vec{n} = 12\hat{i} - 4\hat{j} + 3\hat{k}$. Since the required plane is parallel to $E_{1}$,its equation is of the form $12x - 4y + 3z + k = 0$. Since it passes through $(2, 6, -8)$,we substitute these coordinates: $12(2) - 4(6) + 3(-8) + k = 0$. $24 - 24 - 24 + k = 0 \Rightarrow k = 24$. Thus,the equation is $12x - 4y + 3z + 24 = 0$. Dividing by $12$,we get $x - \frac{y}{3} + \frac{z}{4} + 2 = 0$.

$A$ plane $E_{1}$ makes intercepts $1, -3, 4$ on the coordinate axes. The equation of a plane parallel to plane $E_{1}$ and passing through $(2, 6, -8)$ is

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