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Explain Solution R.D. Sharma Class 12 Chapter 28 The Plane  Exercise 28.13 Question 11 Sub Question 2 Maths Textbook Solution.

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                \begin{aligned} &\vec{n}=3 i-j-k+\lambda(2 i-2 j+k) \end{aligned} is of the form \begin{aligned} p(3+2 \lambda,-1-2 \lambda,-1+\lambda) \\ \end{aligned}  lies in the plane

               \begin{aligned} &\vec{n} \cdot(-9 i-3 j+k) \\ &9(3+2 \lambda)-3(-1-2 \lambda)-(-1+2)=14 \\ &27+18 \lambda-3-6 \lambda+1-\lambda=14 \\ &11 \lambda=-11 \\ &\lambda=-1 \\ \end{aligned}

Thus the required point of intersection is \begin{aligned} &p(3+2 \lambda,-1-2 \lambda,-1+\lambda) \\ \end{aligned} put  value\lambda  in the equation

\begin{aligned} &p[3+2(-1),-1-2(-1),-1+(-1)] \\ &p(1,1,-2) \end{aligned}

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