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|