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#### State True or False for the given statement:The line $r=2\hat{i}-3\hat{j}-\hat{k}+\lambda\left ( \hat{i}-\hat{j}+2\hat{k} \right )$ lies in the plane $r.\left (3\hat{i}+\hat{j}-\hat{k} \right )+2=0$

The equation of the line is given as

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

Any point lying on this line will satisfy the plane equation if the line itself lies in the plane. Also, any point on this line will have a position  vector:

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

Given equation of the plane $\vec{r}.\left ( 3\hat{i}+\hat{j}-\hat{k} \right )+2=0$

If we put a in the above equation,
$\left (\left (2 + \lambda \right )\hat{i}+ \left(-3 - \lambda \right) \hat{j} + \left (-1 + 2\lambda \right)\hat{k} \right ).\left ( 3\hat{i}+\hat{j}-\hat{k} \right ) + 2 \\ = \left (2 + \lambda \right ) \left(3\right) + \left(-3 - \lambda \right) \left(1\right) + \left (-1 + 2\lambda \right)\left(-1 \right) + 2 \\ = 6 - 3\lambda - 3 - \lambda + 1 - 2\lambda + 2 \\ = 5 - 6\lambda \neq R.H.S$

Thus, the line does not lie in the given plane.

Therefore, the given statement is False.