Distance between two parallel planes $$2x + y + 2z = 8$$ and $$4x + 2y + 4z + 5 = 0$$ is

First we recognise that the general form of the equation of a plane is written as $$ax + by + cz + d = 0$$ where the vector $$\vec n = (a,\,b,\,c)$$ is perpendicular (normal) to the plane.

For the first plane we are given the equation

$$2x + y + 2z = 8.$$

To place it in the standard form, we bring all terms to the left‐hand side:

$$2x + y + 2z - 8 = 0.$$

Thus, for this plane we can read off

$$a_1 = 2,\; b_1 = 1,\; c_1 = 2,\; d_1 = -8.$$

The second plane is given as

$$4x + 2y + 4z + 5 = 0.$$

Comparing the normal coefficients of the two planes, we see

$$(4,\,2,\,4) = 2\,(2,\,1,\,2),$$

which means the second plane’s normal vector is exactly twice the first plane’s. Hence the planes are parallel (or possibly coincident) because their normal vectors are proportional.

To use the distance formula conveniently, we first rewrite the second plane so that its normal coefficients match those of the first plane. We do this by dividing every term in the second plane by 2:

$$\frac{4x}{2} + \frac{2y}{2} + \frac{4z}{2} + \frac{5}{2} = 0$$

which simplifies to

$$2x + y + 2z + \frac{5}{2} = 0.$$

Hence in standard form the second plane can be written as

$$2x + y + 2z + \frac52 = 0,$$

so that we now have

$$a_2 = 2,\; b_2 = 1,\; c_2 = 2,\; d_2 = \frac52.$$

Because the coefficients $$a,\,b,\,c$$ of the two planes are now identical, we can apply the distance formula for two parallel planes.

Formula used. If two parallel planes are

$$ax + by + cz + d_1 = 0 \quad\text{and}\quad ax + by + cz + d_2 = 0,$$

then the perpendicular distance $$D$$ between them is

$$D = \frac{\lvert d_2 - d_1 \rvert}{\sqrt{a^2 + b^2 + c^2}}.$$

Now we substitute the values we have just identified:

$$|d_2 - d_1| = \left|\frac52 - (-8)\right| = \left|\frac52 + 8\right|.$$

First convert 8 to a fraction with denominator 2 for easy addition:

$$8 = \frac{16}{2},$$

so

$$\left|\frac52 + 8\right| = \left|\frac52 + \frac{16}{2}\right| = \left|\frac{5 + 16}{2}\right| = \left|\frac{21}{2}\right| = \frac{21}{2}.$$

Next we compute the denominator $$\sqrt{a^2 + b^2 + c^2}$$ with $$a = 2,\; b = 1,\; c = 2$$ :

$$a^2 + b^2 + c^2 = 2^2 + 1^2 + 2^2 = 4 + 1 + 4 = 9,$$

so

$$\sqrt{a^2 + b^2 + c^2} = \sqrt{9} = 3.$$

Putting the numerator and denominator together:

$$D = \frac{\displaystyle \frac{21}{2}}{3} = \frac{21}{2} \times \frac{1}{3} = \frac{21}{6}.$$

We simplify the fraction $$\frac{21}{6}$$ by dividing numerator and denominator by 3:

$$\frac{21}{6} = \frac{21\div3}{6\div3} = \frac{7}{2}.$$

Thus the perpendicular distance between the two given parallel planes is

$$\frac{7}{2}.$$

Hence, the correct answer is Option A.