The equation of the plane passing through the points $(-1,-2,0),(2,3,5)$ and parallel to the line $[\overrightarrow{r}=-3\widehat{j}+\widehat{k}+\lambda (2\widehat{i}+5\widehat{j}-\widehat{k})]$ is
A. $[\overrightarrow{r}\cdot (-30\widehat{i}+13\widehat{j}+5\widehat{k})=4]$
B. $[\overrightarrow{r}\cdot (30\widehat{i}+13\widehat{j}+5\widehat{k})=4]$
C. $[\overrightarrow{r}\cdot (30\widehat{i}+13\widehat{j}-5\widehat{k})=4]$
D. $[\overrightarrow{r}\cdot (30\widehat{i}-13\widehat{j}-5\widehat{k})=4]$

Hint: The equation of the plane passing through any point $[\left( p,q,r \right)]$ is
$[a\left( xp \right)+b\left( yq \right)+c\left( zr \right)=0]$ and
The vector equation of a plane passing through any points $[\left( a,b,c \right)]$ is
$[a\widehat{i}+b\widehat{j}+c\widehat{k}]$
and the equation of plane with the intercept $[\alpha ,\beta ,\And \gamma]$ on the x-axis, y-axis and z-axis respectively is $[x+y+z=1]$




Complete step by step solution:Given, the equation of the plane passes through the points $(-1,-2,0)$. So, the equation of the plane is
$[\begin{align}
  & \Rightarrow a\left( x\left( -1 \right) \right)+b\left( y\left( -2 \right) \right)+c\left( z0 \right)=0 \\
 & \Rightarrow a\left( x+1 \right)+b\left( y+2 \right)+c(z0)=0\text{ }...(1) \\
\end{align}]$
Also, the plane passes through the points $[\left( 2,3,5 \right)]$. So, the equation will be
$[\begin{align}
  & \Rightarrow a\left( 2+1 \right)+b\left( 3+2 \right)+c(50)=0~ \\
 & \Rightarrow 3a+5b+5c=0\text{ }...(2) \\
\end{align}]$
Since the equation of the plane is parallel to the line $[\overrightarrow{r}=-3\widehat{j}+\widehat{k}+\lambda (2\widehat{i}+5\widehat{j}-\widehat{k})]$, the coordinates of the plane are $[\left( 2,5,-1 \right)]$. So, the equation will be
$[2a+5bc=0~...(3)]$
From equations (2) and (3) we get $[{{x}_{1}}=3,\text{ }{{y}_{1}}=5\text{ }{{z}_{1}}=5]$ and $[{{x}_{2}}=2,\text{ }{{y}_{2}}=5\text{ }{{z}_{2}}=-1]$. So, we get
$[\begin{align}
  & \Rightarrow \dfrac{a}{{{y}_{1}}{{z}_{2}}-{{y}_{2}}{{z}_{1}}}=\dfrac{b}{{{z}_{1}}{{x}_{2}}-{{z}_{2}}{{x}_{1}}}=\dfrac{c}{{{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}} \\
 & \Rightarrow ~\dfrac{a}{5\times \left( -1 \right)-5\times 5}=\dfrac{b}{5\times 2-\left( -1 \right)\times 3}=\dfrac{c}{3\times 5-2\times 5} \\
 & \Rightarrow \dfrac{a}{-5-25}=\dfrac{b}{10+3}=\dfrac{c}{15-10} \\
 & \begin{array}{*{35}{l}}
   \Rightarrow \dfrac{a}{-30}=\dfrac{b}{13}=\dfrac{c}{5}=\lambda \\
   a=30\lambda ,~\text{ }b=13\lambda ,~\text{ }c=5\lambda \\
\end{array} \\
\end{align}]$
From equation (1)
$[\begin{align}
  & \Rightarrow 30\lambda \left( x+1 \right)+13\lambda \left( y+2 \right)+5\lambda \left( z0 \right)=0 \\
 & \Rightarrow 30\left( x+1 \right)+13\left( y+2 \right)+5\left( z0 \right)=0 \\
 & \Rightarrow 30x30+13y+26+5z=0 \\
 & \Rightarrow 30x+13y+5z=3026 \\
 & \Rightarrow 30x+13y+5z=4 \\
\end{align}]$
Hence the equation of plane is $[30x+13y+5z=4]$ and vector form of equation is $[\overrightarrow{r}\cdot (-30\widehat{i}+13\widehat{j}+5\widehat{k})=4]$

Thus, Option (A) is correct.

Note: The vector equation of the plane that has three noncollinear points with position vectors $[\widehat{i},\widehat{j},\widehat{k}]$ is $[(\overrightarrow{r}-\widehat{i}).[(\widehat{j}-\widehat{i})times (\widehat{k}-\widehat{i})]=0]$. The direction ratios of the line are proportional to the direction of the cosine line.