Question

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-\alpha z+\beta =0$, then $\left(\alpha ,\beta \right)=$

• A. $\left(6,-17\right)$
• B. $\left(-6,7\right)$
• C. $\left(5,-15\right)$
• D. $\left(-5,15\right)$

Answer 1

The correct option is B

$\left(-6,7\right)$

The explanation for the correct option

The given equation of the line: $L_1:\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$.

The given equation of the plane: $P_1:x+3y-\alpha z+\beta =0$.

The direction ratios of the line $L_1$ can be given by $\left(a_1,b_1,c_1\right)=\left(3,-5,2\right)$.

The direction ratios of the normal to the plane $P_1$ can be given by $\left(a_2,b_2,c_2\right)=\left(1,3,-\alpha \right)$.

As the line $L_1$ lies on the plane $P_1$, thus $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$.

$$\begin{gathered} \Rightarrow \left(3\right)\left(1\right)+\left(-5\right)\left(3\right)+\left(2\right)\left(-\alpha \right)=0 \\ \Rightarrow 3-15-2\alpha =0 \\ \Rightarrow -2\alpha =12 \\ \Rightarrow \alpha =-6 \end{gathered}$$

It is also true as the line $L_1$ lies on the plane $P_1$, thus $\left(2,1,-2\right)$ is on the plane.

So, $x+3y-\alpha z+\beta =0$

$$\begin{gathered} \Rightarrow \left(2\right)+3\left(1\right)-\left(-6\right)\left(-2\right)+\beta =0\left[\because \alpha =-6\right] \\ \Rightarrow 2+3-12+\beta =0 \\ \Rightarrow \beta =7 \end{gathered}$$

Therefore, $\left(\alpha ,\beta \right)=\left(-6,7\right)$.

Hence, (B) is the correct option.