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# Find the distance of the point (-1, -5, -10) from the point of intersection of the line $\hat{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda (3\hat{i} + 4\hat{j} + 2\hat{k})$ and the plane $\hat{r} . (\hat{ i} -\hat{ j} + \hat{k} ) = 5$

Toolbox:
• Distance between two points is $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 1:
The equation of the line is $\overrightarrow r=2\hat i-\hat j+2\hat k+\lambda(3\hat i+4\hat j+2\hat k)$-----(1)
The equation of the plane is $\overrightarrow r.(\hat i-\hat j+\hat k)=5$------(2)
On solving equ(1) and (2)
(i.e)$[2\hat i-\hat j+2\hat k+\lambda(3\hat i+4\hat j+2\hat k)].(\hat i-\hat j+\hat k)=5$
On simplifying we get
$(2\hat i-\hat j+2\hat k).(\hat i-\hat j+\hat k)+\lambda(3\hat i+4\hat j+2\hat k).(\hat i-\hat j+\hat k)=5$
Multiplying by applying dot product we get
$(2+1+2)+\lambda(3-4+2)=5$
Step 2:
On simplifying we get
$5+\lambda=5$
Therefore $\lambda=0$
The point of intersection of the line and the plane is $(2\hat i-\hat j+2\hat k)$
The other point is $(-1,-5,-10)=-\hat i-5\hat j-10\hat k$
Distance between two points is
$\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 3:
Substituting for $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ we get,
$\sqrt {(2-(-1))^2+(-1+5)^2+(2-(-10))^2}$
$\Rightarrow \sqrt{3^2+4^2+12^2}$
$\Rightarrow \sqrt{9+16+144}$
$\Rightarrow \sqrt{169}$
$\Rightarrow 13$
Hence the required distance is $13$