#### Find the vector equation of the line parallel to the vector  $3\hat{i}-2 \hat{j}+3\hat{k}$ and which passes through the point (1, -2, 3).

Given, vector = $3\hat{i}-2\hat{j}+6\hat{k}$

Point = (1, -2, 3)

We can write this point in vector form as $\hat{i}-2\hat{j}+3\hat{k}$

Let ,

$\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$

$\overrightarrow{b}=3\hat{i}-2\hat{j}+6\hat{k}$

We must find the vector equation of the line parallel to the vector $\overrightarrow{b}$ and passing through the point

We know, equation of $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ a line passing through a point and parallel to a given vector is denoted as

Where, $\lambda \epsilon \mathbb{R}$

In other words, we need to find  $\overrightarrow{r}$

This can be achieved by substituting the values of the vectors in the above equation. We get

$\overrightarrow{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$

$\Rightarrow \overrightarrow{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$

This can be further rearranged, upon which we get:

$\Rightarrow \vec{r}=\hat{i}-2\hat{j}+3\hat{k}+3\lambda \hat{j}+3\hat{k}+6\lambda\hat{k}$

$\Rightarrow \vec{r}=\hat{i}+3\lambda \hat{i}-2\hat{j}-2\lambda\hat{j}+3\hat{k}+6\lambda\hat{k}$

$\Rightarrow \vec{r}=\left ( 1-3\lambda \right )\hat{i}+\left ( -2-2\lambda \right )\hat{j}+\left ( 3+6\lambda \right )\hat{k}$

Thus, the require vector equation of line is   $\vec{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$

which can also be written as  $\left ( 1-3\lambda \right )i+\left ( -2-2 \lambda \right )\hat{j}+\left ( 3+6\lambda \right )\hat{k}$