If cartesian equation of the line is \(x-1=2 y+3=3-z\), then its vector equation
is

Given Cartesian equation is
\(x-1 =2 y+3=3-z \)
\( \therefore \frac{x-1}{1} =\frac{2\left(y+\frac{3}{2}\right)}{1}=\frac{-(z-3)}{1} \Rightarrow \frac{x-1}{1}=\) \(\frac{y+\frac{3}{2}}{\left(\frac{1}{2}\right)}=\frac{z-3}{-1}\)
\(\therefore 1, \frac{1}{2},-1 \text { i.e. } 2,1,-2 \text { are d.r. of given line. } \)
\( \therefore \mathrm{A}\left(1,-\frac{3}{2}, 3\right) \text { lies on given line. } \)
\( \therefore \text { vector equation is } \)
\(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}=\left(\hat{\mathrm{i}}-\frac{3}{2} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})\)