Vector Algebra

Given that the vector a is perpendicular to the plane formed by the non-zero vectors b and c, we have:
‌a⋅b=0
‌a⋅c=0
From the problem statement, we know:
|a+b+c|=√|a|2+|b|2+|c|2
Expanding both sides, we get:
|a|2+|b|2+|c|2+2(a⋅b+b⋅c+c⋅a)=|a|2+|b|2+|c|2
This simplifies to:
2(a⋅b+b⋅c+c⋅a)=0
Hence, we conclude:
a⋅b+b⋅c+c⋅a=0
Now, considering |(a×b)⋅c|+|(a×b)×c|, we find:
|(|a||b|sin‌90∘)⋅c|+|(a⋅c)b−(b⋅c)a|
Since sin‌90∘=1, this becomes:
|a||b|c|+0−0‌‌‌ (because ‌a⋅c=0‌ and ‌b⋅c=0)
Thus, the result is:
|a||b||c|