The equation of a plane passing through the points (1, 2, 3), (4, 2, 3) and (-1, 0, 6) is given by:

$-3 y-2 z+12=0$

The correct answer is Option (4) - $-3 y-2 z+12=0$

$\vec a=\hat i+2\hat j+3\hat k$

$\vec b=4\hat i+2\hat j+3\hat k$

$\vec c=-\hat i+0\hat j+6\hat k$

$(\vec r-\vec a).[(\vec b-\vec a)×(\vec c-\vec a)]=0$

$\begin{bmatrix}\vec r-\vec a&\vec b-\vec a&\vec c-\vec a\end{bmatrix}=0$

so $\begin{vmatrix}x-1&y-2&z-3\\3&0&0\\-2&-2&3\end{vmatrix}=0$

$-3\begin{vmatrix}y-2&z-3\\-2&3\end{vmatrix}=0$

$3y-6+2z-6=0$

$-3y-2z+12=0$