\[ \begin{array} ((A)\, 2\; units \quad& (B)\, 4\; units \quad& (C)\, 8\; units \quad& (D) \, \frac{2} { \sqrt{29}} \; units \end{array} \]

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- Distance between parallel plane is $D=\begin{vmatrix}\large\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\end{vmatrix}$

Step 1:

Equation of the planes are :

$2x+3y+4z=4$----(1)

$4x+6y+8z=12$----(2)

Divide equ(2) by 2 through out we get

$2x+3y+4z=6$

Hence this implies the planes are parallel to each other.

Step 2:

We know that the distance between two parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is $ \bigg| \large\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}} \bigg|$

$d_1=6$

$d_2=4$

$\Rightarrow D=\large\frac{6-4}{\sqrt{2^2+3^2+4^2}}$

$\Rightarrow \large\frac{2}{\sqrt{4+9+16}}$

$\Rightarrow D=\large\frac{2}{\sqrt{29}}$

Therefore the distance between two parallel planes =$\large\frac{2}{\sqrt{29}}$units.

Hence D is the correct answer.

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