Question

If the equation of the plane passing through the mirror image of a point $(2,3,1)$with respect to line $\frac{(x+1)}{2}=\frac{(y–3)}{1}=\frac{(z+2)}{-1}$ and containing the line $\frac{(x–2)}{3}=\frac{(1–y)}{2}=\frac{(z+1)}{1}$ is $\alpha x+\beta y+z=24$, then, $\alpha +\beta +\gamma$ is equal to:

• A. $21$
• B. $19$
• C. $18$
• D. $20$

Answer 1

The correct option is B

$19$

Step 1: Solve for the mirror image of the point with respect to the line:

Given $A(2,3,1)$

Let the line $\frac{(x+1)}{2}=\frac{(y–3)}{1}=\frac{(z+2)}{-1}$be $L_1$

We know that if a line is of the form $\frac{x-x_1}{p}=\frac{y-y_1}{q}=\frac{z-z_1}{r}$ then

$\left(x_1,y_1,z_1\right)$ denotes a point on the line and $p,q,r$ denotes the direction cosines.

Let $B$$\left(a,b,c\right)$ is the mirror image of $A$with respect to the line $L_1$

Let $\frac{(x+1)}{2}=\frac{(y–3)}{1}=\frac{(z+2)}{-1}=\lambda$

Then any point on the line $L_1$ is of the form $(2\lambda –1,\lambda +3,–\lambda –2)$

Let point $M$ be the mid point between the point $(2,3,1)$ and its image with respect to the line

Since any point in the line is of the form $(2\lambda –1,\lambda +3,–\lambda –2)$

Therefore, let $M(2\lambda –1,\lambda +3,–\lambda –2)$

Direction ratio's of the line $AM$ are: $\left\{\left(2\lambda –1\right)–2,\left(\lambda +3\right)–3,\left(–\lambda –2\right)–1\right\}$

$=\left\{2\lambda –3,\lambda ,–\lambda –3\right\}$

We know that $AM\perp$$L_1$

Since perpendicular, there the product of the direction ratio will be zero.

$$\begin{gathered} \therefore 2(2\lambda –3)+1\left(\lambda \right)–1(–\lambda –3)=0 \\ \Rightarrow 6\lambda =3 \\ \Rightarrow \lambda =\frac{1}{2} \end{gathered}$$

$\therefore M=(0,\frac{7}{2},–\frac{5}{2})$

$M$ is mid-point of $A\&B$

$\therefore \left(0,\frac{7}{2},-\frac{5}{2}\right)=\left(\frac{a+2}{2},\frac{b+3}{2},\frac{c+1}{2}\right)$

$\therefore \left(a,b,c\right)=\left(-2,4,-6\right)$

Therefore the co-ordinate of $B$ is $\left(-2,4,-6\right)$

Step 2: Solve for the equation of the plane

The plane pass through the point $B$ and contain the line

$\frac{(x–2)}{3}=\frac{(1–y)}{2}=\frac{(z+1)}{1}$

$\frac{(x–2)}{3}=\frac{(y-1)}{-2}=\frac{(z+1)}{1}.........\left(i\right)$

Let the $\frac{(x–2)}{3}=\frac{(y-1)}{-2}=\frac{(z+1)}{1}$ is $L_2$

We know that if the equation of the line is $\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c}$the the line passes through $\left(\alpha ,\beta ,\gamma \right)$ and the direction cosines are $\left(a,b,c\right)$

Point $P$ on line is$(2,1,–1)$

The direction cosine$\left(\overrightarrow{b}\right)$ of the line of line$L_2$ is $\left(3,–2,1\right)$

$\overrightarrow{PB}=\left(-2-2\right)\hat{i}+\left(4-1\right)\hat{j}+\left(-6+1\right)\hat{k}=–4\hat{i}+3\hat{j}–5\hat{k}$ $\left[\because B=\left(-2,4,-6\right)\right]$

We know equation of the plane passing the lines $\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c}$ and $a_1\hat{i}+b_1\hat{j}+c_1\hat{k}$ is

$$\begin{gathered} \left|\begin{array}{ccc}x-\alpha & y-\beta & z-\gamma \\ a& b& c\\ a_1& b_1& c_1\end{array}\right|=0 \\ \therefore \left|\begin{array}{ccc}x-2& y-1& z+1\\ 3& -2& 1\\ 4& -3& 5\end{array}\right|=0 \\ \Rightarrow \left(x-2\right)\left(-10+3\right)+\left(y-1\right)\left(4-15\right)+\left(z+1\right)\left(-9+8\right)=0 \\ \Rightarrow -7x+14-11y+11-z-1=0 \\ \Rightarrow 7x+11y+z=24 \end{gathered}$$

Step 3: Solve for the required values

Now comparing with the equation $\alpha x+\beta y+z=24$ we have

$$\begin{gathered} \therefore \alpha =7\beta =11\gamma =1 \\ \therefore \alpha +\beta +\gamma =19 \end{gathered}$$

Hence, option(B) i.e. $19$ is correct