Question

If for some $\mathrm{\alpha }$ and $\mathrm{\beta }$ in $R$, the intersection of the following three planes $\mathrm{x}+4\mathrm{y}-2\mathrm{z}=1,\mathrm{x}+7\mathrm{y}-5\mathrm{z}=\mathrm{\beta }$ and $\mathrm{x}+5\mathrm{y}+\mathrm{\alpha z}=5$ is a line in $R^3$, then $\mathrm{\alpha }+\mathrm{\beta }$ is equal to:

• A. $0$
• B. $10$
• C. $-10$
• D. $2$

Answer 1

The correct option is B

$10$

Step 1: Given information

$$\begin{gathered} \mathrm{x}+4\mathrm{y}-2\mathrm{z}=1 \\ \mathrm{x}+7\mathrm{y}-5\mathrm{z}=\mathrm{\beta } \\ \mathrm{x}+5\mathrm{y}+\mathrm{\alpha z}=5 \end{gathered}$$

The given plane intersects in a line

$\therefore \mathrm{D}={\mathrm{D}}_{\mathrm{x}}={\mathrm{D}}_{\mathrm{y}}={\mathrm{D}}_{\mathrm{z}}=0$, where is $D$ is the

Step 2: Writing values into matrix form

To find the value of $\mathrm{\alpha }$,

$D=0$

$$\begin{gathered} \therefore \left|\begin{array}{clc}1& 4& -2\\ 1& 7& -5\\ 1& 5& \mathrm{\alpha }\end{array}\right|=0 \\ \Rightarrow 1(7\mathrm{\alpha }+25)-4(\mathrm{\alpha }+5)-2(5-7)=0 \\ \Rightarrow 7\mathrm{\alpha }+25-4\mathrm{\alpha }-20+4=0 \\ \Rightarrow 3\mathrm{\alpha }+9=0 \\ \Rightarrow 3\mathrm{\alpha }=-9 \\ \therefore \mathrm{\alpha }=-3 \end{gathered}$$

Step 3: Finding the value of $\mathrm{\beta }$,

${\mathrm{D}}_{\mathrm{z}}=0$

$$\begin{gathered} \therefore \left|\begin{array}{ccc}1& 4& 1\\ 1& 7& \mathrm{\beta }\\ 1& 5& 5\end{array}\right|=0 \\ \Rightarrow 35-5\mathrm{\beta }-20+4\mathrm{\beta }+5-7=0 \\ \Rightarrow -\mathrm{\beta }+13=0 \\ \Rightarrow -\mathrm{\beta }=-13 \\ \therefore \mathrm{\beta }=13 \end{gathered}$$

Therefore,

$$\begin{gathered} \alpha +\mathrm{\beta }=-3+13=10 \\ \therefore \mathrm{\alpha }+\mathrm{\beta }=10 \end{gathered}$$

Hence, the correct answer is an option (B).