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), the \alpha is equal to :
Option: 1 10
Option: 2 -10
Option: 3 5
Option: 4 -5

Answers (1)

\\\text { Hence normal is } \perp^{\mathrm{r}} \text { 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}})

\begin{array}{l} \vec{n}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 2 & 3 & -1 \end{array}\right|=\hat{i}(2-6)-\hat{j}(-1-4)+\hat{k}(3+4) \\ \vec{n}=-4 \hat{i}+5 \hat{j}+7 \hat{k} \end{array}

\begin{aligned} &\text { Now equation of plane passing through }(3,1,1) \text { is }\\ &\Rightarrow-4(x-3)+5(y-1)+7(z-1)=0\\ &\Rightarrow-4 x+12+5 y-5+7 z-7=0\\ &\Rightarrow-4 x+5 y+7 z=0 \end{aligned}

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

\begin{array}{l} -4 \alpha+5 \times(-3)+7 \times(5)=0 \\ \Rightarrow-4 \alpha-15+35=0 \\ \Rightarrow \alpha=5 \end{array}

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