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If $\bar{r}=(2-\lambda+\mu) \bar{i}+(1-\mu) \bar{j}+(2-3 \lambda+2 \mu) \bar{k}$ is the vector equation of a plane, then the equivalent cartesian equation of the plane is
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Verified Answer
The correct answer is:
$3 x+y-z=5$
$$
\begin{aligned}
& \text {Given } \bar{\gamma}=(2-\lambda+\mu) \hat{\mathrm{i}}+(1-\mu) \hat{\mathrm{j}}+(2-3 \lambda+2 \mu) \hat{\mathrm{k}} \\
& \bar{\gamma}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+(-\hat{\mathrm{i}}-3 \hat{\mathrm{k}}) \lambda+(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \mu
\end{aligned}
$$
is a plane passing through $(2 \hat{i}+\hat{j}+2 \hat{k})$ parallel to $(-\hat{i}-3 \hat{k})$ and $(\hat{i}-\hat{j}+2 \hat{k})$
$$
\begin{aligned}
& \operatorname{now}(-\hat{\mathrm{i}}-3 \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\
& \Rightarrow \bar{\eta}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}
\end{aligned}
$$
now equation of the plane is $\bar{\gamma} \cdot \bar{\eta}=\overline{\mathrm{d}}$
$$
\Rightarrow 3 \mathrm{x}+\mathrm{y}-\mathrm{z}=5
$$
\begin{aligned}
& \text {Given } \bar{\gamma}=(2-\lambda+\mu) \hat{\mathrm{i}}+(1-\mu) \hat{\mathrm{j}}+(2-3 \lambda+2 \mu) \hat{\mathrm{k}} \\
& \bar{\gamma}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+(-\hat{\mathrm{i}}-3 \hat{\mathrm{k}}) \lambda+(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \mu
\end{aligned}
$$
is a plane passing through $(2 \hat{i}+\hat{j}+2 \hat{k})$ parallel to $(-\hat{i}-3 \hat{k})$ and $(\hat{i}-\hat{j}+2 \hat{k})$
$$
\begin{aligned}
& \operatorname{now}(-\hat{\mathrm{i}}-3 \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\
& \Rightarrow \bar{\eta}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}
\end{aligned}
$$
now equation of the plane is $\bar{\gamma} \cdot \bar{\eta}=\overline{\mathrm{d}}$
$$
\Rightarrow 3 \mathrm{x}+\mathrm{y}-\mathrm{z}=5
$$
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