# Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.9 Question 11 maths textbook solution.

Answer :  The answer of the given question is $\sqrt{29}$ Units

Hint :

$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$

Given :

3 Point $(7,2,4)$ and the plane determine by the point

$A(2,5,-3), B(-2,-3,5) \&(5,3,-3)$

Solution :

The equation of the plane passing through $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right) \&\left(x_{3}, y_{3}, z_{3}\right)$ is given by the following equation

$\left|\begin{array}{lll} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0$

According to questions

\begin{aligned} &\left(x_{1}, y_{1}, z_{1}\right)=(2,5,-3) \\ &\left(x_{2}, y_{2}, z_{2}\right)=(-2,-3,5) \\ &\left(x_{3}, y_{3}, z_{3}\right)=(5,3,-3) \end{aligned}

Putting these values

$\left|\begin{array}{ccc} x-2 & y-5 & z-(-3) \\ -4 & -8 & 8 \\ 3 & -2 & 0 \end{array}\right|=0$

\begin{aligned} &(x-2)(16)+(y-5)(24)+(z+3)(32)=0 \\ &2 x+3 y+4 z-7=0 \end{aligned}

Distance of $2 x+3 y+4 z-7=0$ from $(7,2,4)$

We know, the distance of point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane

$\pi: a x+b y+c z+d=0$ Is given by

\begin{aligned} &P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ &P=\left|\frac{2 \times 7+3 \times 2+4 \times 4-7}{\sqrt{2^{2}+3^{2}+4^{2}}}\right| \end{aligned}

$P= \sqrt{29}$ Units

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