# Find the vector equation of the plane determined by the points $A\left ( 3,-1,2 \right ),B\left ( 5,2,4 \right )\: and\: C\left ( -1,-1,6 \right )$. Hence, find the distance of the plane, thus obtained, from the origin.

Equation of plane determined by $A\left ( 3,-1,2 \right )\: B\left ( 5,2,4 \right )\: and\: C\left ( -1,-1,6 \right )$ is
$\begin{vmatrix} x-3 &y+1 &x-2 \\ 5-3 &2+1 &4-2 \\ -1-3& -1+1 &6-2 \end{vmatrix}= 0$
$\Rightarrow \begin{vmatrix} x-3 &y+1 &x-2 \\ 2 &3 &2 \\ -4& 0 &4 \end{vmatrix}= 0$
$= 12\left ( x-3 \right )-16\left ( y+1 \right )+12\left ( z-2 \right )= 0$
$\Rightarrow 3x-4y+3z-19= 0$
so vector eq is $\vec{r}.\left ( \frac{ 3\hat{i} -4\hat{j}+3\hat{k}}{\sqrt{9+16+9}} \right )=\frac{19}{\sqrt{34}}\, i.e\, \vec{r}\cdot \left ( \frac{ 3\hat{i} -4\hat{j}+3\hat{k}}{\sqrt{34}} \right )= \frac{19}{\sqrt{34}}$
$\therefore$  the distance of the plane (i) from origin is $= \frac{19}{\sqrt{34}}$

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