Q75 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics
If $\bar{a} = \hat{i} + \hat{j} + \hat{k}, \bar{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\bar{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k}$ are three vectors in which $\bar{c}$ lies in the plane of $\bar{a}$ and $\bar{b}$, then $x = \cdots$
- A$0$
- B$1$
- C$-4$
- Correct answerD$-2$
Solution
Since $\bar{c}$ lies in the plane of $\bar{a}$ and $\bar{b}$, the vectors $\bar{a}$, $\bar{b}$, and $\bar{c}$ are coplanar.
Therefore, their scalar triple product is zero: $[\bar{a} \quad \bar{b} \quad \bar{c}] = 0$
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x - 2 & -1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((-1)(-1) - 2(x - 2)) - 1(1(-1) - 2x) + 1(1(x - 2) - (-1)x) = 0$
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$
$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$
$4 + 2x = 0$
$2x = -4 \Rightarrow x = -2$
Answer: $-2$
Therefore, their scalar triple product is zero: $[\bar{a} \quad \bar{b} \quad \bar{c}] = 0$
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x - 2 & -1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((-1)(-1) - 2(x - 2)) - 1(1(-1) - 2x) + 1(1(x - 2) - (-1)x) = 0$
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$
$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$
$4 + 2x = 0$
$2x = -4 \Rightarrow x = -2$
Answer: $-2$
Hence, the correct answer is option: 'D'.