$\begin{array}{1 1}(A) x+2y-3z-4=0 \\(B) x+2y+3z-14=0 \\(C) x+2y-3z-14=0 \\ (D) x+2y-3z+14=0 \end{array} $

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- Direction ratios of a given vector is whose points are $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is $(x_2-x_1),(y_2-y_1),(z_2-z_1)$.
- Equation of the plane is $a(x-x_1)+b(y_2-y_1)+c(z_2-z_1)=0$

Step 1:

Let the points $o$ be $(0,0,0)$ and $P(1,2,-3)$

Therefore direction ratios of $OP$ are

$(x_2-x_1),(y_2-y_1),(z_2-z_1)$

(i.e)$(1-0),(2-0),(-3-0)$

$\Rightarrow (1,2,-3)$

Step 2:

The plane passing through $(x_1,y_1,z_1)$ is

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

Where $a,b,c$ are the direction ratio of normal are $(1,2,-3)$ and the point $P$ is $(1,2,3)$.

Therefore the equation of the required plane is $1(x-1)+2(y-2)+3(z-3)=0$

$x-1+2y-4-3z-9=0$

On simplifying

$x+2y-3z-14=0$

Hence equation of required plane is $x+2y-3z-14=0$

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