Find the equation of the plane through the intersection of the planes
3x-y+2z-4 =0 and x+y+z-2=0 and the point (2, 2, 1).
The equation of any plane through the intersection of the planes,
3x-y+2z-4=0 and x+y+z-2=0, is
(3x-y+2z-4)+ \(\alpha\) (x+y+z-2)=0, where \(\alpha \in\) R...(1)
The plane passes through the point (2 ,2,1).
Therefore, this point will satisfy equation(1).
∴ (3×2-2+2×1-4)+α(2+2+1-2)=0
\(\Rightarrow \) 2+3\(\alpha\) =0
\(\Rightarrow \alpha=-\frac{2}{3}\)
Substituting \(\alpha=-\frac{2}{3}\) in equation(1), we obtain
(3x-y+2z-4)-\(\frac{2}{3}\) (x+y+z-2)=0
\(\Rightarrow 3\) (3x-y+2z-4)-2(x+y+z-2)=0
\(\Rightarrow\) (9x-3y+6z-12)-2(x+y+z-2)=0
\(\Rightarrow \) 7x-5y+4z-8=0
This is the required equation of the plane.
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}


A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely: