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The problem is as follows: $$ \left\{ \begin{array}{c} x^2 + y^2 + z^2 = \frac{45}4 \\ (x-1)^2 + (y-1)^2 + (z-1)^2 = \frac{45}4 \\ x + 2y + z - 4 = 0 \end{array} \right. $$

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    Next time, show your **own** work. – 2017-02-28
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    okey, as for the problem I actually did the same as the people below but I miss-put some signs so it didn't come well, now Ive seen where I was wrong. Next time I'm going to do it right. Thanks – 2017-02-28

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from the second equation we get $$x^2+y^2+z^2-2(x+y+z)=\frac{45}{4}-3$$ using the first equation with zhis obove we obtain $$\frac{45}{4}-2(x+y+z)=\frac{45}{4}-3$$ from here we get $$x+y+z=\frac{3}{2}$$ and $$x+2y+z=4$$ can you get further?

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    I've got it, thanks a lot! – 2017-02-28
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Expanding equation (2),

$x^2-2x+1+y^2-2y+1+z^2-2z+1=\frac{45}4$

$\frac{45}4 - 2x - 2y - 2z + 3 = \frac{45}4$

$2x + 2y + 2z = 3$

$x+y+z=\frac{3}2$

Using this equation and equation (3) you can eliminate x and z to find y. Then find x and z.