The equation of a line passing through a point \(\vec{a}\) and parallel to a vector \(\vec{b}\) is given by \(\vec{r} = \vec{a} + t\vec{b}\), where \(t\) is a scalar.
Here, the point is \((4, -3, 7)\), so \(\vec{a} = 4\hat{i} - 3\hat{j} + 7\hat{k}\).
The vector parallel to the line is \(3\hat{i} + \hat{j} + 2\hat{k}\), so \(\vec{b} = 3\hat{i} + \hat{j} + 2\hat{k}\).
Therefore, the equation of the line is \(\vec{r} = (4\hat{i} - 3\hat{j} + 7\hat{k}) + t(3\hat{i} + \hat{j} + 2\hat{k})\).
In component form, this is:
\(x = 4 + 3t\)
\(y = -3 + t\)
\(z = 7 + 2t\)
Rearranging, we get:
\(x = 3t + 4\)
\(y = t - 3\)
\(z = 2t + 7\)
Correct Answer: x=3t+4, y=t-3, z=2t+7
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