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If $v_{1}=(1,1,0)^{T}, v_{2}=(0,1,1)^{T}, v_{3}=(1,1,1)^{T}$

And $Proj_{S}(v_{3})=a_{1}v_{1}+a_{2}v_{2}$, where $a_{1},a_{2}$ are scalars, $S$ is spanned by $v_{1}$ and $v_{2}$

Does $a_{1}=a_{2}=\frac{2}{3}$?

This is the equation I used: enter image description here

Also does anyone know how to check this with WolframAlpha? I can't seem to figure it out.

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    Didn't you forget something? Say, to tell what $\;S\;$ is...? – 2017-01-20
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    Oh S is spanned by v1 and v2. Sorry it's been quite a while since I took linear algebra and the actual interpretation of this stuff is quite hazy to me. – 2017-01-20
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    Add this relevant piece of info in your question. – 2017-01-20
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    $v_3-Proj_{S}(v_3)$ is perpendicular to $v_1$ and $v_2$ by definition. This will let you compute a and b. – 2017-01-20

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By definition (or by what can be proved from what I think the standard definitions are), we get:

$$S:=\text{Span}\,\{v_1,v_2\}\implies \text{Proj}_S\,v_3:=\frac{\langle v_3,v_1\rangle}{\left\|v_1\right\|^2}\,v_1+\frac{\langle v_3,v_2\rangle}{\left\|v_2\right\|^2}\,v_2$$

and in your case

$$\text{Proj}_S\,v_3:=\frac22\begin{pmatrix}1\\1\\0\end{pmatrix}+\frac22\begin{pmatrix}0\\1\\1\end{pmatrix}=\ldots$$