Q.Find the distance between the planes $x - 2y + 2z = 6$ and $3x - 6y + 6z = 2$.

Write both planes with the same normal vector, then use the parallel-planes distance formula $d=\dfrac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$.

Given planes: $x+y+3z=4$ and $2x+2y+6z=10$.

Divide the second equation by 2:

$x+y+3z=5$

Both planes now have the same normal vector $(1,1,3)$, confirming they are parallel:

$x+y+3z=4 \quad\text{and}\quad x+y+3z=5$

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write both planes with the same normal vector, then use the parallel-planes distance formula

Planes: $2x+3y+6z+5=0$ and $2x+3y+6z-9=0$ (rewriting $2x+3y+6z=9$).

Both share the normal vector $(2,3,6)$, $|\vec n|=\sqrt{4+9+36}=\sqrt{49}=7$.

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