Q71 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics
A point lying on the plane that passes through the points $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ is
- A$-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
- B$-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$
- Correct answerC$\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$
- D$4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
Solution
Equation of plane passing through 3 given point is
$$
\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|=0
$$
So,
$$
\left|\begin{array}{ccc}
x-1 & y+1 & z-1 \\
0 & -1 & 2 \\
0 & 3 & -4
\end{array}\right|=0
$$
$$
\begin{aligned}
(x-1)(4-6)-(y+1)(0)+(z-1)(0) & =0 \\
\Rightarrow \quad-2 x+2 & =0
\end{aligned}
$$
So, $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ satisfy the first equation $(1,1,-1)$
$$
\Rightarrow \quad-2(1)+2=0
$$
Hence, the correct answer is option: 'C'.