A plane passing through the point (3,1,1)


A plane passing through the point $(3,1,1)$ contains two lines whose direction ratios are 1 , $-2,2$ and $2,3,-1$ respectively. If this plane also passes through the point $(\alpha,-3,5)$, then $\alpha$ is equal to:

  1. -10

  2. 5

  3. 10

  4. -5

Correct Option: , 2


Hence normal is $\perp^{\mathrm{r}}$ to both the lines so normal vector to the plane is

$\overrightarrow{\mathrm{n}}=(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})$

$\overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -2 & 2 \\ 2 & 3 & -1\end{array}\right|=\hat{\mathrm{i}}(2-6)-\hat{\mathrm{j}}(-1-4)+\hat{\mathrm{k}}(3+4)$

$\overrightarrow{\mathrm{n}}=-4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$

Now equation of plane passing through $(3,1,1)$ is


$\Rightarrow-4 x+12+5 y-5+7 z-7=0$

$\Rightarrow-4 x+5 y+7 z=0$......(1)

Plane is also passing through $(\alpha,-3,5)$ so this point satisfies the equation of plane so put in equation (1)

$-4 \alpha+5 \times(-3)+7 \times(5)=0$

$\Rightarrow-4 \alpha-15+35=0$

$\Rightarrow \alpha=5$


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