JEE Main 2022 — Mathematics Vectors & 3D Geometry

Given lines, $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$0x−2​=1y−1​=1z​ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$2x−3​=2y−5​=1z−1​

Now direction ratio of line of shortest distance will be,

$$|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 2 & 2 & 1\end{matrix}|=-\hat{i}+2\hat{j}-2\hat{k}$$∣i^02​j^​12​k^11​∣=−i^+2j^​−2k^

Now given angle between line and plane is ${\mathrm{sin}}^{-1}\sqrt{\frac{2}{27}}=\alpha$sin−1272​​=α, so $\mathrm{sin}\alpha =\sqrt{\frac{2}{27}}$sinα=272​​

Now direction ratio of normal to plane is $(a,-1,-1)$(a,−1,−1)

So, $\mathrm{sin}\alpha =|\frac{-a-2+2}{\sqrt{4+4+1}\sqrt{{a}^{2}+1+1}}|=\frac{\sqrt{2}}{3\sqrt{3}}$sinα=∣4+4+1​a2+1+1​−a−2+2​∣=33​2​​

$$\Rightarrow \sqrt{3}|a|=\sqrt{2}\sqrt{{a}^{2}+2}$$⇒3​∣a∣=2​a2+2​

$$\Rightarrow 3{a}^{2}=2{a}^{2}+4$$⇒3a2=2a2+4

$$\Rightarrow {a}^{2}=4$$⇒a2=4

$$\Rightarrow a=2(\text{as}a>0\text{given})$$⇒a=2(asa>0given)

Now equation of plane will be $2x-y-z=0$2x−y−z=0 and image of point $(1,1,-5)$(1,1,−5) will be,

$$\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z+5}{-1}=-2\frac{(2-1+5)}{{2}^{2}+{1}^{2}+{1}^{2}}$$2x−1​=−1y−1​=−1z+5​=−222+12+12(2−1+5)​

$$\Rightarrow \frac{x-1}{2}=\frac{y-1}{-1}=\frac{z+5}{-1}=-2$$⇒2x−1​=−1y−1​=−1z+5​=−2

$$\Rightarrow x=-3,y=3,z=-3$$⇒x=−3,y=3,z=−3

So, $\alpha =-3,\beta =3,\gamma =-3$α=−3,β=3,γ=−3

So, $\alpha +\beta -\gamma =-3+3+3=3$α+β−γ=−3+3+3=3

Note: This question was bonus in JEE mains 25th july 2022 paper, we have done some modification to the original question.