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Question : 31 of 160
Marks:
+1,
-0
Solution:
|a|=‌3|b|=7|c|=‌13∣a+‌b+‌+c|2=|a|2+|b|2+|c|2‌+2(a⋅b+b⋅c+c⋅a)=‌9+49+169+2((6−6−4)‌+(−24+9−24)+(−4−6+24)=‌227+2(−4−39+14)=‌227−58=‌169⇒|a+b+c|=13⇒√|a|+|b|+|c|+|a+b+c|=‌√10+13+13=‌6
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