Question

Let a line with direction ratios $a,–4a,–7$ be perpendicular to the lines with direction ratios $3,–1,2b$ and $b,a,–2$. If the point of intersection of the line $\frac{x+1}{a^2+b^2}=\frac{y-2}{a^2-b^2}=\frac{z}{1}$ and the plane $x–y+z=0$ is $(\alpha ,\beta ,\gamma )$, then $\alpha +\beta +\gamma$ is equal to.

Answer 1

Given $a\cdot 3+(–4a)(–1)+(–7)2b=0.....\left(1\right)$

and $ab–4a^2+14=0...\left(2\right)$

$\Rightarrow a^2=4\text{and}{b}^2=1$

$\therefore \text{Line}L=\frac{x+1}{5}=\frac{y-2}{3}=\frac{Z}{1}=\lambda \text{(say)}$

$\Rightarrow \text{General point on line is}(5\lambda –1,3\lambda +2,\lambda )$ for finding point of intersection with $x–y+z=0$
we get $(5\lambda –1)–(3\lambda +2)+\left(\lambda \right)=0$

$\Rightarrow 3\lambda -3=0\Rightarrow \lambda =1$

$\therefore$ Point at intersection $(4,5,1)$

$\therefore \alpha +\beta +\gamma =4+5+1=10$