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Q)

The point of intersection of the line $\overrightarrow{r}=(\overrightarrow{i}-\overrightarrow{k}) + t(\overrightarrow{3i}+\overrightarrow{2j}+\overrightarrow{7k})$ and the plane $\overrightarrow{r}. (\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})=8 $ is

\[\begin{array}{1 1}(1)(8 , 6 , 22 )&(2)(-8 ,-6 ,-22 ) \\(3)(4 , 3 , 11) &(4) (-4 , -3 , -11 )\end{array}\]

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A)
$\overrightarrow{r}=(\overrightarrow{i}-\overrightarrow{k}) + t(\overrightarrow{3i}+\overrightarrow{2j}+\overrightarrow{7k})$
Put $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j} +z\overrightarrow{k}$
$x\overrightarrow{i}+y\overrightarrow{j} +z\overrightarrow{k}=(\overrightarrow{i}-\overrightarrow{k})+t(3 \overrightarrow{i}+2 \overrightarrow{j}+\overrightarrow{k})$
Equating the $\overrightarrow{i},\overrightarrow{j},\overrightarrow{k}$ terms on the both sides
$x=1+3t,y=2t,z=-1+7t$
$\large\frac{x-1}{3}$$=t$
$\large\frac{y}{2}$$=t$
$\large\frac{z+1}{7}=t$
The cartesian equation of the plane is
$\overrightarrow{r}.(\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})=8$
$(x\overrightarrow{i}+y\overrightarrow{j} +z\overrightarrow{k}).(\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})=8$
$x+y+z=8$
(-8,-6,-22) is the only point satisfying these two equations
Hence 2 is the correct answer.
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