Answer
$12x+4y+3z=12$
Work Step by Step
The plane containing the point $P(p,q,r)$ and having the normal vector $n=\lt a,b,c\gt$ is expressed algebraically by the equation as follows:
$a(x-p)+b(y-q)+c(z-r)=0$
Let $a,,b,c$ be the x-,y-,z- intercepts respectively.
Now we have when the line passes through x-axis at $1$,this means that the x-intercept is $1$.
So, $a=1$
when the line passes through y-axis at $3$,this means that the y-intercept is $3$.
So, $b=3$
when the line passes through z-axis at $4$,this means that the z-intercept is $4$.
So, $c=4$
The intercept form is given as: $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$
or, $\dfrac{x}{1}+\dfrac{y}{3}+\dfrac{z}{4}=1$
Thus, we have $12x+4y+3z=12$
