Find the intercepts cut off by the plane 2x+y-z = 5
2x+y-z=5...(1)
Dividing both sides of equation(1) by 5, we obtain
\(\frac{2}{5}x+\frac{y}{5}-\frac{z}{5}=1\)
\(\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1\)...(2)
It is known that the equation of a plane in intercept form is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\), where a,b,c are the intercepts cut off by the planet at x, y, and z axes respectively.
Therefore, for the given equation, a=\(\frac{5}{2}\), b=5, and c=-5
Thus, the intercepts cut off by the plane are \(\frac{5}{2}\), 5, and -5.
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: