Take derivative and set to 0:

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• \|\vec{d}(t)\|^2 = (2 - 2t)^2 + (-1 - t)^2 + (5 + t)^2
• Add them up:
• \|\vec{d}(t)\|^2 = (4 - 8t + 4t^2) + (1 + 2t + t^2) + (25 + 10t + t^2) = 6t^2 + 4t + 30
• \frac{d}{dt}(6t^2 + 4t + 30) = 12t + 4 = 0 \Rightarrow t = -\frac{1}{3}
• Substitute $t = -\frac{1}{3}$ into $\vec{r}(t)$:
• \vec{r}\left(-\frac{1}{3}\right) = \begin{pmatrix} 1 + 2(-\frac{1}{3}) \\ -\frac{1}{3} \\ -3 - (-\frac{1}{3}) \end{pmatrix} = \begin{pmatrix} 1 - \frac{2}{3} \\ -\frac{1}{3} \\ -3 + \frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ -\frac{1}{3} \\ -\frac{8}{3} \end{pmatrix}

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\[ 2a = 1 \]\[ a = \frac{1}{2} \]Substitute \(a = \frac{1}{2}\) into Equation 4:\[ 3\left(\frac{1}{2}\right) + b = 1 \]\[ \frac{3}{2} + b = 1 \]\[ b = 1 - \frac{3}{2} = -\frac{1}{2} \]Substitute \(a = \frac{1}{2}\) and \(b = -\frac{1}{2}\) into Equation 1:\[ \frac{1}{2} - \frac{1}{2} + c = 2 \]\[ c = 2 \]Thus, the coefficients are \(a = \frac{1}{2}\), \(b = -\frac{1}{2}\), and \(c = 2\).