Search any question & find its solution
Question:
Answered & Verified by Expert
If $a, b, c$ are real numbers such that $a+b+c=0$ and $a^{2}+b^{2}+c^{2}=1$, then $(3 a+5 b-8 c)^{2}+(-8 a+3 b+5 c)^{2}+(5 a-8 b+3 c)^{2}$ is equal to
Options:
Solution:
1960 Upvotes
Verified Answer
The correct answer is:
147
Expanding are get
$\begin{array}{l}
98\left(a^{2}+b^{2}+c^{2}\right)-98(a b+b c+c a) \\
\quad=98-98\left(-\frac{1}{2}\right) \\
\quad=147
\end{array}$
$\begin{array}{l}
98\left(a^{2}+b^{2}+c^{2}\right)-98(a b+b c+c a) \\
\quad=98-98\left(-\frac{1}{2}\right) \\
\quad=147
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.