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  4. If the equation of the plane passing through the point (-1,2,0) and parallel to the lines (x/3)=(y+1/0)=(z-2/-1) and (x-1/1)=(y+1/2)=(z+1/-1) is a x+b y+c z=1, then the value of (a+b+c), is

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Q. If the equation of the plane passing through the point $(-1,2,0)$ and parallel to the lines $\frac{x}{3}=\frac{y+1}{0}=\frac{z-2}{-1}$ and $\frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-1}$ is $a x+b y+c z=1$, then the value of $(a+b+c)$, is

Vector Algebra

Solution:

Normal vector of the plane $h=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -1 \\ 1 & 2 & -1\end{vmatrix}$
$h=2 \hat{ i }+2 \hat{ j }+6 \hat{ k }=2(\hat{ i }+\hat{ j }+3 \hat{ k })$
$\therefore$ Equation of plane $1( x +1)+1( y -2)+3( z -0)=0$
P: $x + y +3 z =1$
hence, $(a+b+c)=1+1+3=5$.