JEE 2023 — Mathematics PYQ

Equation of plane P containing the given lines is

$(x+2y+3z-4)+\lambda (2x+y-z+5)=0\Rightarrow (1+2\lambda )x+(2+\lambda )y+(3-\lambda )z+(-4+5\lambda )=0$

Now, plane $P$ is perpendicular to plane $P^{\prime }$

$\vec{r}=(\hat{i}-\hat{j})+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (-\hat{i}+2\hat{j}+3\hat{k})$

So, normal to plane $P^{\prime }$ is

$\vec{n}=(\hat{i}+\hat{j}+\hat{k})\times (\hat{i}-2\hat{j}+3\hat{k})$

$\Rightarrow \vec{n}=5\hat{i}-2\hat{j}-3\hat{k}$

∴ P and P' are perpendicular

∴ $5(1+2\lambda )-2(2+\lambda )-3(3-\lambda )=0$

∴ $5+10\lambda -4-2\lambda -9+3\lambda =0$

∴ $11\lambda =8\Rightarrow \lambda =\frac{8}{11}$

$\therefore P:\left(1+\frac{16}{11}\right)x+\left(2+\frac{8}{11}\right)y+\left(3-\frac{8}{11}\right)z+\left(5\times \frac{8}{11}-4\right)=0$

i.e., $27x+30y+25z=4$

∴ $a=27,b=30$ and $c=25$

∴ $a-b+c=27-30+25=22$