$\text{Position vector of four points}A,B,C,D\text{are}-\hat{i}+\hat{j}+\hat{k},3\hat{i}-2\hat{j}+2\hat{k},4\hat{i}-\lambda \hat{j}-\hat{k}\text{and}\hat{i}+\hat{j}+\hat{k}\text{ respectively.}$

$\text{The value of}\mathrm{ \lambda }\text{ for which the points}\mathrm{ A},B,C,D\text{are coplanar is}$

The correct option is -7.

To find the value of $\lambda$ for which the four points $A,B,C,D$ are coplanar, we can represent their position vectors as follows:

$\overrightarrow{\mathrm{OA}}=-\hat{i}+\hat{j}+\hat{k}$

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

$\overrightarrow{\mathrm{OC}}=4\hat{i}-\lambda \hat{j}-\hat{k}$

$\overrightarrow{\mathrm{OD}}=\hat{i}+\hat{j}+\hat{k}$

Four points $A,B,C,D$ are coplanar if the vectors $\overrightarrow{\mathrm{AB}}$, $\overrightarrow{\mathrm{AC}}$, and $\overrightarrow{\mathrm{AD}}$ lie in the same plane. This condition is met when their scalar triple product is zero:
$\left[\overrightarrow{\mathrm{AB}}\overrightarrow{\mathrm{AC}}\overrightarrow{\mathrm{AD}}\right]=0$

Let us first find the components of the vectors $\overrightarrow{\mathrm{AB}}$, $\overrightarrow{\mathrm{AC}}$, and $\overrightarrow{\mathrm{AD}}$:

1. $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(3-(-1\left)\right)\hat{i}+(-2-1)\hat{j}+(2-1)\hat{k}=4\hat{i}-3\hat{j}+\hat{k}$

2. $\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(4-(-1\left)\right)\hat{i}+(-\lambda -1)\hat{j}+(-1-1)\hat{k}=5\hat{i}-(\lambda +1)\hat{j}-2\hat{k}$

3. $\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OA}}=(1-(-1\left)\right)\hat{i}+(1-1)\hat{j}+(1-1)\hat{k}=2\hat{i}+0\hat{j}+0\hat{k}$

Now, we set the determinant of the matrix formed by these three vectors to zero:

$\left|\begin{array}{ccc}4& -3& 1\\ 5& -(\lambda +1)& -2\\ 2& 0& 0\end{array}\right|=0$

To compute the determinant, we expand along the third row (since it has two zero elements):

$2·\left[\right(-3)·(-2)-1·(-(\lambda +1)\left)\right]=0$

Simplify the terms inside the brackets:

$2·[6+(\lambda +1\left)\right]=0$

$2·(\lambda +7)=0$

Divide by 2 to solve for $\lambda$:

$\lambda +7=0$

$\lambda =-7$

Thus, the value of $\lambda$ for which the points are coplanar is -7.