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The plane through the intersection of planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel to Y-axis also passes through the point
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Verified Answer
The correct answer is:
$(3,2,1)$
Equation of plane through the intersection of given planes is
$\begin{aligned}
& (x+y+z-1)+\lambda(2 x+3 y-z+4)=0 \quad \ldots \text { (i) } \\
& \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+4 \lambda-1=0
\end{aligned}$
Since the plane is parallel to $\mathrm{Y}$-axis.
$\begin{aligned}
\therefore \quad & 1+3 \lambda=0 \\
& \Rightarrow \lambda=\frac{-1}{3}
\end{aligned}$
Substituting $\lambda=\frac{-1}{3}$ in (i), we get
$\begin{aligned}
& (x+y+z-1)-\frac{1}{3}(2 x+3 y-z+4)=0 \\
& \Rightarrow x+4 z-7=0
\end{aligned}$
Point $(3,2,1)$ satisfies this equation.
$\begin{aligned}
& (x+y+z-1)+\lambda(2 x+3 y-z+4)=0 \quad \ldots \text { (i) } \\
& \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+4 \lambda-1=0
\end{aligned}$
Since the plane is parallel to $\mathrm{Y}$-axis.
$\begin{aligned}
\therefore \quad & 1+3 \lambda=0 \\
& \Rightarrow \lambda=\frac{-1}{3}
\end{aligned}$
Substituting $\lambda=\frac{-1}{3}$ in (i), we get
$\begin{aligned}
& (x+y+z-1)-\frac{1}{3}(2 x+3 y-z+4)=0 \\
& \Rightarrow x+4 z-7=0
\end{aligned}$
Point $(3,2,1)$ satisfies this equation.
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