Question

If $x^2+xy+xz=135,y^2+yz+xy=351$ and $z^2+xz+yz=243$, then $x^2+y^2+z^2=?$

• A. $300$
• B. $275$
• C. $250$
• D. $225$

Answer 1

The correct option is B $275$

$x^2+xy+xz=\mathrm{135........}\left(i\right)y^2+yz+xy=\mathrm{351........}\left(ii\right)z^2+xz+yz=\mathrm{243........}\left(iii\right)$

Adding (i), (ii), (iii), we get,

$x^2+y^2+z^2+2(xy+yz+xz)=729=>{(x+y+z)}^2={27}^{2}x+y+z=\mathrm{27........}\left(iv\right)$

In equation (i),

$x^2+xy+xz=135=>x(x+y+z)=135=>x\times 27=135=>x=5$

Similarly in equation (ii), we get,

$y^2+yz+xy=351=>y(x+y+z)=351=>y\times 27=351=>y=13$

Again in equation (iii), we get,

$z^2+xz+yz=243=>z(x+y+z)=243=>z\times 27=243=>z=9$

$x^2+y^2+z^2=5^2+{13}^2+9^2=275$