121366 Let \((\alpha, \beta, \gamma)\) be the image of the point \(P(2,3,5)\) in the plane \(2 x+y-3 z=6\). Then \(\alpha+\beta+\gamma\) is equal to

121368 Let the plane \(P\) pass through the intersection of the planes \(2 x+3 y-z=2\) and \(x+2 y+3 z=\) 6 , and be perpendicular to the plane \(2 x+y-z\) \(+1=0\). If \(d\) is the distance of \(P\) from the point \((-7,1,1)\), then \(d^2\) is equal to

121370 Let \(Q\) be the mirror image of the point \(P(1,2,1)\) with respect to the plane \(x+2 y+2 z=16\). Let \(T\) be a plane passing through the point \(Q\) and contains
the
line \(\overrightarrow{\mathbf{r}}=-\hat{\mathbf{k}}+\lambda(\hat{\mathbf{1}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}), \lambda \in \mathbb{R}\). Then, which of the following points lies on \(T\) ?

121371 If the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}\) and \(\frac{x-a}{1}=\frac{y+2}{2}=\frac{z-3}{1}\) intersect at the point \(P\), then the distance of the point \(P\) from the plane \(\mathrm{z}=\mathbf{a}\) is :

121366 Let \((\alpha, \beta, \gamma)\) be the image of the point \(P(2,3,5)\) in the plane \(2 x+y-3 z=6\). Then \(\alpha+\beta+\gamma\) is equal to

121368 Let the plane \(P\) pass through the intersection of the planes \(2 x+3 y-z=2\) and \(x+2 y+3 z=\) 6 , and be perpendicular to the plane \(2 x+y-z\) \(+1=0\). If \(d\) is the distance of \(P\) from the point \((-7,1,1)\), then \(d^2\) is equal to

121370 Let \(Q\) be the mirror image of the point \(P(1,2,1)\) with respect to the plane \(x+2 y+2 z=16\). Let \(T\) be a plane passing through the point \(Q\) and contains
the
line \(\overrightarrow{\mathbf{r}}=-\hat{\mathbf{k}}+\lambda(\hat{\mathbf{1}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}), \lambda \in \mathbb{R}\). Then, which of the following points lies on \(T\) ?

121371 If the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}\) and \(\frac{x-a}{1}=\frac{y+2}{2}=\frac{z-3}{1}\) intersect at the point \(P\), then the distance of the point \(P\) from the plane \(\mathrm{z}=\mathbf{a}\) is :

121366 Let \((\alpha, \beta, \gamma)\) be the image of the point \(P(2,3,5)\) in the plane \(2 x+y-3 z=6\). Then \(\alpha+\beta+\gamma\) is equal to

121368 Let the plane \(P\) pass through the intersection of the planes \(2 x+3 y-z=2\) and \(x+2 y+3 z=\) 6 , and be perpendicular to the plane \(2 x+y-z\) \(+1=0\). If \(d\) is the distance of \(P\) from the point \((-7,1,1)\), then \(d^2\) is equal to

121370 Let \(Q\) be the mirror image of the point \(P(1,2,1)\) with respect to the plane \(x+2 y+2 z=16\). Let \(T\) be a plane passing through the point \(Q\) and contains
the
line \(\overrightarrow{\mathbf{r}}=-\hat{\mathbf{k}}+\lambda(\hat{\mathbf{1}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}), \lambda \in \mathbb{R}\). Then, which of the following points lies on \(T\) ?

121371 If the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}\) and \(\frac{x-a}{1}=\frac{y+2}{2}=\frac{z-3}{1}\) intersect at the point \(P\), then the distance of the point \(P\) from the plane \(\mathrm{z}=\mathbf{a}\) is :

121366 Let \((\alpha, \beta, \gamma)\) be the image of the point \(P(2,3,5)\) in the plane \(2 x+y-3 z=6\). Then \(\alpha+\beta+\gamma\) is equal to

121368 Let the plane \(P\) pass through the intersection of the planes \(2 x+3 y-z=2\) and \(x+2 y+3 z=\) 6 , and be perpendicular to the plane \(2 x+y-z\) \(+1=0\). If \(d\) is the distance of \(P\) from the point \((-7,1,1)\), then \(d^2\) is equal to

121370 Let \(Q\) be the mirror image of the point \(P(1,2,1)\) with respect to the plane \(x+2 y+2 z=16\). Let \(T\) be a plane passing through the point \(Q\) and contains
the
line \(\overrightarrow{\mathbf{r}}=-\hat{\mathbf{k}}+\lambda(\hat{\mathbf{1}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}), \lambda \in \mathbb{R}\). Then, which of the following points lies on \(T\) ?

121371 If the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}\) and \(\frac{x-a}{1}=\frac{y+2}{2}=\frac{z-3}{1}\) intersect at the point \(P\), then the distance of the point \(P\) from the plane \(\mathrm{z}=\mathbf{a}\) is :