Question

If the straight lines $\frac{\mathrm{x}-1}{\mathrm{k}}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-3}{3}\text{and}\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-3}{\mathrm{k}}=\frac{\mathrm{z}-1}{3}$intersect at a point, then the integer $k$ is equal to:

• A. $-2$
• B. $-5$
• C. $5$
• D. $2$

Answer 1

The correct option is B

$-5$

Explanation of correct option.

Let ${\mathrm{L}}_1:\frac{\mathrm{x}-1}{\mathrm{k}}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-3}{3}$ is the equation of the line passing through the point $\left(1,2,3\right)$.

And let ${\mathrm{L}}_2:\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-3}{\mathrm{k}}=\frac{\mathrm{z}-1}{3}$is the equation of the line passing through the point $\left(2,3,1\right)$.

Let ${\mathrm{L}}_3$ be a line passing through the points $\left(1,2,3\right)$ and $\left(2,3,1\right)$.

So, the direction cosines of ${\mathrm{L}}_3$ is given by $\left(2-1,3-2,1-3\right)\equiv \left(1,1,-2\right)$

As ${\mathrm{L}}_3$ passes through $\left(1,2,3\right)$ and $\left(2,3,1\right)$ and ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$ intersect at a point.

Therefore, ${\mathrm{L}}_1$ , ${\mathrm{L}}_2$ and ${\mathrm{L}}_3$ are coplanar

Therefore,

$$\begin{gathered} \left|\begin{array}{ccc}k& 2& 3\\ 3& k& 2\\ 1& 1& -2\end{array}\right|=0 \\ \Rightarrow 2k^2+5k-25=0 \\ \Rightarrow k=\frac{5}{2},-5 \end{gathered}$$

Hence, Option (B) is correct.