NDA Math Three Dimensional Geometry

We have to find the angle between the planes 2x−y+z=1 and x+y+2z=3
So, a1=2,b1=−1,c1=1 and a2=1,b2=1 and c2=2
Let θ will be angle between these two plane
‌ So, ‌cos‌θ=‌

a1a2+b1b2+c1c2

√(a1)2+(b1)2+(c1)2√(a2)2+(b2)2+(c2)2

=‌

2×1+(−1)×1+1×2

√22+(−1)2+12√12+12+22

⇒cos‌θ=‌

2−1+2

√6√6

=‌

3

6

=‌

1

2

=cos‌

π

3

∴θ=π∕3
So, statement 1 is true.
Now, we have to the find distance between the planes
6x−3y+6z+2=0‌ and ‌
2x−y+2z+4=0
Rewrite the equation 6x−3y+6z+2=0
⇒3(2x−y+2z+2∕3)=0
⇒2x−y+2z+2∕3=0
So, 2x−y+2z+2∕3=0 and 2x−y+2z+4=0 are parallel
a=2,b=−1,c=2,d1=2∕3‌ and ‌d2=4
‌ Distance ‌=|‌

d1−d2

√a2+b2+c2

|=|‌

‌ 2 3 −4

√22+−12+22

|=|‌

‌ −10 3

√9

|=‌

10

3×3

=‌

10

9

So, statement 2 is true.