15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T51e,

Given lines are $$L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$ and $$L_2: \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$$

$$\Rightarrow$$ Direction ratios of $$L_1:< 2,3,4>$$

<>$$\Rightarrow$$ Direction ratios of $$L_2:<3,4,5>$$

Since, plane contains these two lines, so normal to plane $$\| \overrightarrow{L_1} \times \overrightarrow{L_2}$$

$$\begin{aligned} \mathbf{L}_1 \times \mathbf{L}_2 & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right| \\ & =\hat{\mathbf{i}}(15-16)-\hat{\mathbf{j}}(10-12)+\hat{\mathbf{k}}(8-9) \\ & =-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned}$$

$$\therefore$$ Direction ratios of normal to plane are $$<-1,2,-1>$$.

$$L$$, passes through $$(1,2,3)$$.

So, plane also passes through $$(1,2,3)$$

$$\therefore$$ Its equation is

$$(-1)(x-1)+2(y-2)+(-1)(z-3)=0$$

i.e., $$x-2 y+z=0$$

Other parallel plane is $$\alpha x-2 y+z-k=0$$

Since, they are parallel

$$\therefore \frac{\alpha}{1}=\frac{-2}{-2}=\frac{1}{1} \Rightarrow \alpha=1$$

Distance between these planes $$=\sqrt{6}$$ units

$$\Rightarrow \frac{|0-(-k)|}{\sqrt{\left(1^2\right)+(-2)^2+(1)^2}}=\sqrt{6} \Rightarrow|k|=6$$

Three Dimensional Geometry

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