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#### . If the line drawn from the point (-2, -1, -3) meets a plane at right angle at the point (1, -3, 3), find the equation of the plane.

Given: the line drawn from point (-2, -1, -3) meets a plane at 900 at the point (1, -3, 3). We must find the equation of the plane.

Any line perpendicular to the plane is the normal.

Let the points be P (-2, -1, -3) and Q (1, -3, 3), then the line PQ is a normal to the plane.

Hence, PQ = (1 + 2, -3 + 1, 3 + 3)=> PQ = (3, -2, 6)

=> Normal to the plane = $\vec{PQ}$

$\vec{PQ}=3\hat{i}-2\hat{j}+6\hat{k}$

The vector equation of a plane is represented by $\left (\vec{r}-\vec{a} \right ).\vec{n}=0$

Putting the obtained values in this equation,

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\ \vec{a}=\hat{i}-3\hat{j}+3\hat{k}\\ \vec{n}=3\hat{i}-2\hat{j}+6\hat{k}$

We get,

$\Rightarrow \left (\left (x\hat{i}+y\hat{j}+z\hat{k} \right )-\left (\hat{i}-3\hat{j}+3\hat{k} \right ) \right ).\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )=0\\ \Rightarrow \left ( x\hat{i}-y\hat{j}+z\hat{k}-\hat{i}+3\hat{j}-3\hat{k} \right ).\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )=0\\ \Rightarrow \left ( x\hat{i}-\hat{i}+y\hat{j}+3\hat{j}+z\hat{k}-3\hat{k} \right ).\left (3\hat{i}-2\hat{j}+6\hat{k} \right )=0\\ \Rightarrow \left ((x-1)\hat{i}+(y+3)\hat{j}+(z-3)\hat{k} \right ).\left (3\hat{i}-2\hat{j}+6\hat{k} \right )=0$

$\Rightarrow 3(x - 1) + (-2)(y + 3) + 6(z - 3) = 0\\ \Rightarrow3(x - 1) - 2(y + 3) + 6(z - 3) = 0\\ \Rightarrow 3x- 3 -2y - 6 + 6z - 18 = 0\\ \Rightarrow 3x - 2y +6z - 3 - 6 - 18 = 0\\ \Rightarrow 3x - 2y + 6z - 9 - 18 = 0\\ \Rightarrow 3x - 2y + 6z - 27 = 0\\ \Rightarrow 3x - 2y + 6z = 27$

Therefore, the required equation of the plane is 3x - 2y + 6z = 27.