Question
Determine the point on yz-plane which is equidistant from points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1).

Answer

The coordiante of x point on yz-plane is 0
Let the point be P(0, y, z).
Now, PA = PB
$\sqrt{(2-0)^2+(0-\text{y})^2+(3-\text{z})^2}=\sqrt{(0-0)^2+(3-\text{y})^2+(2-\text{z})^2}$
$\Rightarrow4+\text{y}^2+9-6\text{z+}\text{z}^2=9-6\text{y}+\text{y}^2+4-4\text{z}+\text{z}^2$
$\Rightarrow-6\text{z}=-6\text{y}-4\text{z}$
$\Rightarrow3\text{y}-\text{z}=0\ ...\text{(i)}$
Also, PA = PC
$\sqrt{(2-0)^2+(0-\text{y})^2+(3-\text{z})^2}=\sqrt{(0-0)^2+(0-\text{y})^2+(1-\text{z})^2}$
$\Rightarrow4+\text{y}^2+9-6\text{z+}\text{z}^2=\text{y}^2+1-2\text{z}+\text{z}^2$
$\Rightarrow13-6\text{z}=1-2\text{z}$
$\Rightarrow-4\text{z}=-12$
$\Rightarrow\text{z}=3\ ...\text{(ii)}$
Solving (i) and (ii), we get
y = 1
Hence, the coordinates of the point is (0, 1, 3).

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