Chi Squared Clarification

$$ \left[ \begin{array}{c} \frac{Z_1-2Z_2+Z_3}{\sqrt{12}} \\[4pt] \frac{Z_1-Z_3} 2 \\[6pt] \frac{Z_1-Z_2} 2 \\[6pt] \frac{Z_1+Z_2-2Z_3}{\sqrt{12}} \end{array} \right] = \left[ \begin{array}{rrr} 1/\sqrt{12} & -2/\sqrt{12} & 1/\sqrt{12} \\ 1/2 & 0 & -1/2 \\ 1/2 & -1/2 & 0 \\ 1/\sqrt{12} & 1/\sqrt{12} & -2/\sqrt{12} \end{array} \right] \left[ \begin{array}{c} Z_1 \\ Z_2 \\ Z_3 \end{array} \right] \tag1 $$ Observe that the rank of this $4\times3$ matrix is only $2,$ i.e. you have only two linearly independent columns, since the second column is $-1$ times the sum of the first and third columns.

Let us find two unit vectors orthogonal to each other that span the column space of that matrix: $$ \left[ \begin{array}{c} -\sqrt3/\sqrt8 \\ +1/\sqrt8 \\ -1/\sqrt8 \\ +\sqrt3/\sqrt8 \end{array} \right], \qquad \left[ \begin{array}{c} +1/\sqrt8 \\ +\sqrt3/\sqrt8 \\ +\sqrt3/\sqrt8 \\ +1/\sqrt8 \end{array} \right] $$ Write the column vector on the left side of line $(1)$ above as a linear combination of these two vectors, each of the coefficients $U_1,U_2$ being a linear combination of $Z_1,Z_2,Z_3.$

Then show that $\operatorname{var}(U_1)= \operatorname{var}(U_2)=1$ and $\operatorname{cov}(U_1,U_2)=0$ and that $U_1,U_2\sim\operatorname N(0,1),$ and that $U_1^2 + U_2^2$ is equal to the sum of squares that you say should have a $\chi^2_2$ distribution.