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(C) The equation of a line passing through a point $\vec{a} = x_1 \bar{i} + y_1 \bar{j} + z_1 \bar{k}$ and parallel to a vector $\vec{b} = a \bar{i} +

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$\frac{(x-1)}{1}=\frac{(y+2)}{1}=\frac{(z-1)}{3}$

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(C) The equation of a line passing through a point $\vec{a} = x_1 \bar{i} + y_1 \bar{j} + z_1 \bar{k}$ and parallel to a vector $\vec{b} = a \bar{i} + b \bar{j} + c \bar{k}$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Given point is $(1, -2, 1)$,so $x_1 = 1, y_1 = -2, z_1 = 1$.Given parallel vector is $\bar{i} + \bar{j} + 3 \bar{k}$,so $a = 1, b = 1, c = 3$.Substituting these values into the formula,we get $\frac{x-1}{1} = \frac{y-(-2)}{1} = \frac{z-1}{3}$.This simplifies to $\frac{x-1}{1} = \frac{y+2}{1} = \frac{z-1}{3}$.

The Cartesian equation of the line passing through the point $\bar{i}-2 \bar{j}+\bar{k}$ and parallel to the vector $\bar{i}+\bar{j}+3 \bar{k}$ is

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