Let the direction of the line be $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ . ∴ The equation of the line through (1, 2, 3) having the direction $\vec{b}$ is $\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda\vec{b}$ $\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})$ $+\lambda(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})$ ...(1) Since line (1) is parallel to $\vec{r}\cdot(\hat{i}-\hat{j}+2\hat{k})=5$ $(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})\cdot(\hat{i}-\hat{j}+2\hat{k})=0$ $\Rightarrow b_1 - b_2 + 2b_3 = 0$ ...(2) Again since line (1) is parallel to $\vec{r}\cdot(3\hat{i}+\hat{j}+\hat{k})=6$ $\therefore(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})\cdot(3\hat{i}+\hat{j}+\hat{k})=0$ $\Rightarrow 3b_1 + b_2 + b_3 = 0$ ...(3)Solving (2) and (3), we get $\frac{b_1}{-3}=\frac{b_2}{5}=\frac{b_3}{4}$∴< – 3, 5, 4> are direction ratios of the line. ∴ The required equation of the line is $\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})$ $+\lambda(-3\hat{i}+5\hat{j}+4\hat{k})$