The Cartesian equation of a line is 2x - 3 = | vedclass

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(C) The given Cartesian equation is $2x - 3 = 3y + 1 = 5 - 6z$. To write this in standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}

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$r = (7 \hat{i} - 5 \hat{j}) + \lambda(3 \hat{i} + 2 \hat{j} - \hat{k})$

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(C) The given Cartesian equation is $2x - 3 = 3y + 1 = 5 - 6z$. To write this in standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$,we divide by the coefficients of $x, y, z$: $2(x - \frac{3}{2}) = 3(y + \frac{1}{3}) = -6(z - \frac{5}{6})$. Dividing by $6$,we get $\frac{x - 3/2}{3} = \frac{y + 1/3}{2} = \frac{z - 5/6}{-1}$. The direction vector of this line is $\vec{v} = 3 \hat{i} + 2 \hat{j} - \hat{k}$. The vector equation of a line passing through point $\vec{a} = 7 \hat{i} - 5 \hat{j} + 0 \hat{k}$ and parallel to $\vec{v}$ is given by $\vec{r} = \vec{a} + \lambda \vec{v}$. Thus,$\vec{r} = (7 \hat{i} - 5 \hat{j}) + \lambda(3 \hat{i} + 2 \hat{j} - \hat{k})$.

The Cartesian equation of a line is $2x - 3 = 3y + 1 = 5 - 6z$. The vector equation of the line passing through the point $(7, -5, 0)$ and parallel to the given line is

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