Question

Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z–7a=0=2x–5y–z–3,aϵR$ at the point
$P(\alpha ,\beta ,\gamma )$. Then the value of $\alpha +\beta +\gamma$ equals _____.

• A. 12
• B. 12.00
• C. 12.0

Answer 1

Answer: (A), (B), (C)

Equation of plane containing the line
$4ax–y+5z–7a=0=2x–5y–z–3$ can be written as
$4ax–y+5z–7a+\lambda (2x–5y–z–3)=0$
$(4a+2\lambda )x–(1+5\lambda )y+(5–\lambda )z–(7a+3\lambda )=0$
Which is coplanar with the line

$\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$

$4(4a+2\lambda )+(1+5\lambda )–(7a+3\lambda )=0$
$9a+10\lambda +1=0$ …(1)
and
$(4a+2\lambda )1+(1+5\lambda )2+5–\lambda =0$
$4a+11\lambda +7=0$ …(2)
$a=1,\lambda =–1$
Equation of plane is $x+2y+3z–2=0$
Intersection with the line

$\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$= $t$

$(7t+3)+2(–t+2)+3(–4t+3)–2=0$
$–7t+14=0$
$t=2$
So, the required point is $(17,0,–5)$
$\alpha +\beta +\gamma =12$