The equation $\frac{3 x + 1}{2} = \frac{y - 1}{2} = \frac{- z + 5}{3}$ represents a straight line then the vector form of this line will be

\to r = - \frac{1}{3} i + j + 5 k + t (\frac{3}{2} i + 2 j - 3 k)

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\to r = \frac{3}{1} i + j + 5 k - t (\frac{2}{3} i + 2 j - 3 k)

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\to r = - \frac{1}{3} i + j + 5 k + t (\frac{2}{3} i + 2 j - 3 k)

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Solution

The correct option is C $\to r = - \frac{1}{3} i + j + 5 k + t (\frac{2}{3} i + 2 j - 3 k)$
given equation is

$\frac{3 x + 1}{2} = \frac{y - 1}{2} = \frac{- z + 5}{3}$

$\frac{x + \frac{1}{3}}{\frac{2}{3}} = \frac{y - 1}{2} = \frac{z - 5}{- 3}$

vector form of above equation is

$\to r = - \frac{1}{3}^i +^j + 5^k + t (\frac{2}{3}^i + 2^j - 3^k)$

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\vec{a} = \hat{i} - \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} - \hat{k}, \vec{c} = λ \hat{i} - \hat{j} + λ \hat{k}

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